WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to be talking about solids that are not solids of revolution.
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Let us jump right on in.
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So far, we have been talking about regions by taking a given function.
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Let me write all of this down.
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We have taken regions in the xy plane bounded by given functions and rotated them around a given axis.
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That is what we have been doing, rotated them around a given axis.
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Doing, we generated these things called solids of revolution.
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We can take something like this.
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We will take the xy axis and we will take a function, let us just say y = √x, something like that.
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And then, we rotate it, in this case, let us say around the x axis.
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We rotate this axis and we end up generating this solid like that.
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What we do is we end up actually… I have drawn it slightly off so that you can actually see the rotation part.
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Let me actually erase that.
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Let me just draw it fully sideways.
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Just a straight, it goes like that, something like that.
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What we did is we took a slice of it.
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That slice, when we turn this slice this way, what you end up getting is a circle.
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You get a circle, it is a solid disk.
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This is what we have been doing so far.
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We have exploited the circular quality that comes from revolution around an axis.
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Any rotation is going to generate some thing where the cross section is going to be a circle.
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The area of the circle is very easy, it is just π r².
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We exploited that axis to find the cross sectional area.
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In other words, the area of the cross section.
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Take a cross section which is always going to be perpendicular to the axis of rotation.
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We turn this way and the figure that we get, that is the cross section.
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Let us say that again, a cross section is obtained by slicing a solid perpendicular to a given axis.
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In this case, we sliced perpendicular to the axis of rotation.
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In the case above, area is equal to π r².
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Our other types of objects, where they are not quite solid,
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let us say you have this function and this function, and we rotate it around the x axis.
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You are going to generate a different type of solid.
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Not completely solid, it is kind of empty.
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In this case, when we rotate it, what you end up generating is,
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Let me do this in red.
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If we take a slice and a slice, what you end up with is not a full circle but you end up with something which is a washer.
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It is the same thing, it is still an area.
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In this case, we are still exploiting the circular quality.
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Now the area is just π × the outer radius - the inner radius², that is it.
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The area of the outer circle which is π r, outer radius².
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Let me actually write it out.
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It is going to be π × the outer radius² - π × the inner radius².
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The area = π × the outer radius² - the inner radius².
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The squared is on the outside, I apologize for that.
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In each case, we added up all of the slices.
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We added up which is integration.
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We integrated the volumes for each slice.
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That is it, that is all we did.
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We found the volume of one slice and then we integrated it.
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We added up all of the volumes.
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In each case, we added up the volumes for each slice whether it was a disk or a washer.
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The volume is just the integral from some beginning point to some endpoint.
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In other words, some beginning point to some endpoint of the area which is a function of x dx.
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Or if we were integrating along the y axis, the area which is a function of y dy.
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That is it, that is all we have been doing.
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Find the area and you integrate.
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I will write it out.
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In each case, the cross section was circular.
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The question is, what happens if the cross section is not circular?
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What if we have a cross section which is not circular?
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The general formula is still the same.
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The general formula still holds.
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In other words, the volume, our total volume is going to equal the integral from a to b of some area function,
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whatever the area of that cross sectional shape is now.
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In the case of rotation around an axis, we get a circle.
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It does not have to be rotation, there can be another cross section.
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You can have a square, triangle, trapezoid, whatever shape, but it is still just the area × the little width of the slice.
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That does not change.
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Our job is to find this, the area as a function of x or y depending on which way you are integrating.
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That is really all we are doing, ay dy.
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And then, you add up all of the individual volumes.
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ax or ay, depending on our choice, ay are still cross sectional areas but of various shapes.
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We have area formulas for all the different shapes, the basic shapes in geometry that we deal with, it is not a problem.
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We must use our knowledge which we got from pre-calculus, geometry, algebra,
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whatever it is, to derive a formula for a(x) or a(y).
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That is really all we are doing, that is the hardest part of these problems.
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It is kind of like the max/min problem or the related rates problem.
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The hard part is not the calculus, the hard part is coming up with a formula
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from the information that you are given in the physical situation.
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You must use our knowledge to derive a formula for a, x, y, then just integrate.
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And that part at this point should not cause any problem.
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The integration, that should be the least of our worries.
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It is the least of our worries, we want to find an integral, that is we want to do.
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Let us go ahead and do an example, and see what we can do with that.
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Find the volume of a pyramid whose base is a square of side length s and whose height is h.
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Let us draw this out in perspective.
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Let us go ahead and draw a square base, put this up there.
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We have this pyramid, like that.
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Side length s, we will call that s.
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Let me draw a fully straight on view of this.
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This is going to be not a perspective drawing.
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Side length is s and this is h.
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What we are going to do is were going to take a slice of this.
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We are going to take a slice which is a square.
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We are going to take the area of the square.
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We are going to multiply by dx.
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We are going to find that, that is going to be the volume of that one slice.
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We are going to find the volume of the slice which is like that, if we are looking straight on.
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And then, we are just going to add up all of the slices.
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That is it, that is all we are doing.
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Except in this case, the cross sectional area is a square, it is not a circle.
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You just need to arrange the problem, in such a way on your coordinate axis.
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That will give you the answer that you want.
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We are going to take slices, find the area of the squares.
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Find the volumes of the slices, and we are going to integrate.
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In other words, add up.
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That is it, that is what we are going to do.
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Let us set this problem up.
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I’m going to set it up this way.
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There is more than one way to do this.
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Please do not think that this is the way to do it.
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You are going to arrange it in a way that makes the most sense to you,
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based on the coordinate system or what it is that your particular mind sees.
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Clearly, you have discovered by now, even back in pre-calculus,
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that as these problems become more sophisticated and more complicated, there is more than one approach to the problem.
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What your approach works, by all means, use it.
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I'm going to set up my coordinate system in such a way.
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I’m going to take this pyramid and I’m going to turn it on its side.
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I'm going to run the vertical axis along the x axis because I want to integrate along x.
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I’m going take this.
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When we are looking at side view of this thing, it is going to look something like that.
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This right here, that is our s, that is our side length.
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This right here, that is our h.
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The slice that we are going to take is this slice right here.
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That is our slice.
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This slice, I’m going to rotate it this way.
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This is a slice that we are taking, perpendicular to an axis that runs through the center of the pyramid.
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This, when I turn this around, the square that I'm looking at, the slice that I'm looking at is going to be that.
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The axis, this axis, passes right through that.
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It is right through the middle.
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This thing right here, I’m just going to go ahead and call this y.
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This is s and I’m going to call it m for the time being.
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M, because it is not really s, it is a different smaller size than s.
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I’m just going to call it m for the time being.
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The area of this slice is equal to m².
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We want to find m², in terms of the things they gave us, s, h, and x,
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because that is the variable that I’m integrating along.
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I’m going to be adding up all the slices this way.
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That is what I'm doing.
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We have a relationship here.
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Let me work in red here.
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If this value is x, this value is going to be y.
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This is my y and this is my x.
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There is a relationship here.
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x is related to y, as h is related to this length right here which is s/2.
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I set up a proportion.
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This implies that x s/2 is equal to hy.
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I’m going to solve for y here because I'm looking for y.
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I’m looking for y so that I can multiply it by 2 and have them be my m.
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Then, I’m going to square that m and that is going to be my area.
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y is equal to x × s/ 2h.
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m is equal to 2y which is equal to 2x s/ 2h.
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The 2's cancel and I'm left with xs/h, that is my m.
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I found my m, that is just x × s/h.
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The area we said is equal to m², that is equal to xs/h², that is equal to s²/ h² x².
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That is the area of our square.
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Now I'm going to multiply it by the differential.
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This width is dx, it is my old differential width.
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The volume of each slice = s²/ h² x² dx.
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Now I add them all up.
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The total volume = the integral from, we had 0.
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This length was h, 0 to h s²/ h² x² dx.
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You are just adding up all the different slices.
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Volume = the integral from 0 to h.
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S²/ h² is a constant so it becomes s²/ h² × the integral from 0 to h of x² dx.
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The rest is easy.
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s²/ h², x³/ 3 from 0 to h.
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We put in h, we should get s²/ h² × h³/ 3 – 0.
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We are left with the final answer of s² × h/3.
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Side length s, height h, the volume is s² h/3.
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1/3 size² × the height, which you already knew back from geometry days.
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Calculus is how we actually derive that formula that we use in geometry.
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That is it, nice and straightforward.
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You just need to set it up properly, in a way that makes sense to you and just sort of figure out the rest.
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Set up the coordinate system, take a slice, find the area of a cross section of that slice, and then integrate.
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I think that was the last of that one.
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Example 2, a certain solid has a circular base of radius 2, cross sectional areas perpendicular to the base are equilateral triangles.
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What is the volume of the solid?
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The hardest part of this problem, all of these problems, is visualizing what is going on.
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Just follow what is says, draw it out, and everything should fall out.
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Let me go back the black here.
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I have got this, the certain solid has a circular base of radius 2.
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I have got a circular base and my radius is 2.
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This is 2, this is -2, this is 2, and this is -2.
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Cross sectional area is perpendicular to the base.
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If this is my base, perpendicular means hit it that way.
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In other words, I’m going right down into the base.
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If I just drop something straight down into this, cross sectional area perpendicular to the base.
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When I take a slice like that, are equilateral triangles.
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Cross sectional areas, I hit the base perpendicularly and I turned that around.
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Basically what is happening here now, they are equilateral triangles.
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This actually looks like this.
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When I take this and I turn around, I get something that looks like this.
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If I call this point A and this point B, this is A and this is B.
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We are going to rotate it so I look at it a little bit better.
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Do I really need to rotate it so I look at it a little bit better?
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This axis point is right there.
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This axis is right there, all I have done is hit it, turn it around.
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Now I'm looking at this figure.
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When I rotate, I’m going to bring this just because I'm used to looking at triangles this way.
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This is A, now this is B, my axis is here.
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That is it, we need to find the area of the triangle, the volume, in terms of,
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in this particular case, this is my y axis, this is my x axis.
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My slice is this way, I'm going to be integrating along the x axis.
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I need to find some function of x.
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We will be integrating along the x axis.
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We need an area function for the triangle as a function of x.
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In other words, we need some a(x).
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Let us see what we can do.
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Let me draw it again.
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I have my circle, I have my slice.
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I turned it, rotated it.
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Now that I’m looking at a triangle that looks like this, equilateral triangle.
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Here is my axis.
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If this is x value, this is my y value.
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That y value is this, that is my y value.
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That is my height, it is equilateral.
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This angle is 60°.
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I know the fundamental formula for the area of the triangle.
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It is equal to base × height/ 2.
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Let us see what the relationship is here, between the x and y.
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This is a circle of radius 2, I have got x² + y² = 2.
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I have got y² = 2 - x².
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Therefore, y is equal to √2 - x², that is y.
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The base is twice y.
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The base is equal to twice y, the base is equal to 2 × √2 - x².
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I found my base, what about my height?
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My height, this is a 60° triangle.
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This is 30, 60, 90.
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If this is y and height is just y√3.
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It is just y√3, the height is equal to √3 × √2 - x².
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I have got my height and I have got my base.
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Plug them into this equation.
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The area is equal to base × height divided by 2.
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It is equal to 2 × √2 - x² × √3 × √2 - x², that is h/2.
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My 2’s cancel, this × that.
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I’m left with an area function.
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My area at some function of x is going to equal √3 × 2 - x².
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That is my area, that is my area of the triangle.
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When I look at it from that end, that is my area.
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I’m going to find the volume and I’m just going to add up everything,
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then I’m going to integrate from -2 to 2 because I’m integrating along the x axis.
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We have our area function which is equal to √3 × 2 - x².
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My differential volume element is just √3 × 2 - x² × dx.
00:28:05.000 --> 00:28:13.400
My total volume is equal to the integral from, once again, we have our circle.
00:28:13.400 --> 00:28:15.200
This is -2, this is 2.
00:28:15.200 --> 00:28:21.000
Our slice is here, we are adding up all the slices this way, along the x axis.
00:28:21.000 --> 00:28:33.700
It is going to be from -2 to 2 of the area function or of this thing right here, √3 × 2 - x² dx.
00:28:33.700 --> 00:28:38.900
That is it, that is my integral, this is what I was seeking.
00:28:38.900 --> 00:28:50.900
The volume is equal to √3 comes out, from -2 to 2 of 2 - x² dx.
00:28:50.900 --> 00:28:54.100
In the problems that I have written now, I actually went ahead and I solved this integral.
00:28:54.100 --> 00:28:56.400
But I think at this point, I hope you will forgive me.
00:28:56.400 --> 00:29:00.400
I’m just going to go ahead and leave these integrals to you because they are easy enough to solve.
00:29:00.400 --> 00:29:11.200
This is just going to be √3 × 2x – x³/ 3 taken from -2 to 2.
00:29:11.200 --> 00:29:12.300
I will leave that to you.
00:29:12.300 --> 00:29:13.800
The difficult part was this.
00:29:13.800 --> 00:29:17.900
This is what we wanted, this is what is important.
00:29:17.900 --> 00:29:24.900
Finding the integral, the rest is just techniques of integration, whatever function you happen to be dealing with.
00:29:24.900 --> 00:29:29.000
That is example 2, let us see what we have got.
00:29:29.000 --> 00:29:36.300
Find the volume of a pyramid whose base is an equilateral triangle of side length a and whose height is h.
00:29:36.300 --> 00:29:38.600
Let us go ahead and draw a perspective of this.
00:29:38.600 --> 00:29:41.200
Let us go ahead and do this in black first.
00:29:41.200 --> 00:29:47.000
Our perspective diagram is going to be something like this.
00:29:47.000 --> 00:29:56.000
I will put that there, I will put a little point here.
00:29:56.000 --> 00:30:02.100
This is side a, this is side a.
00:30:02.100 --> 00:30:07.000
This is side a, we have this pyramid whose base is an equilateral triangle.
00:30:07.000 --> 00:30:09.700
The base is an equilateral triangle.
00:30:09.700 --> 00:30:14.300
I will go ahead and draw it straight on.
00:30:14.300 --> 00:30:17.300
We know what we are looking at.
00:30:17.300 --> 00:30:21.700
We have this little triangle, I’m going to draw it along,
00:30:21.700 --> 00:30:25.700
We are looking at it this way, this edge.
00:30:25.700 --> 00:30:33.800
From your perspective, the triangle is coming out like that.
00:30:33.800 --> 00:30:41.300
This length is a and the height is h.
00:30:41.300 --> 00:30:44.900
Not altogether different than the first example that we did.
00:30:44.900 --> 00:30:54.700
This is our perspective view and this is straight on.
00:30:54.700 --> 00:30:57.100
Let us go ahead and draw this out.
00:30:57.100 --> 00:31:00.600
Again, I’m going to use the same coordinate system.
00:31:00.600 --> 00:31:03.200
I’m going to set it up like this.
00:31:03.200 --> 00:31:13.600
I'm going to have it so that the axis, my x axis again runs right through the center of the triangle.
00:31:13.600 --> 00:31:19.100
I’m basically going to take this thing, I’m just going to turn it this way onto the axis.
00:31:19.100 --> 00:31:32.500
There is that, there is this.
00:31:32.500 --> 00:31:41.200
There is that, I’m going to take a slice.
00:31:41.200 --> 00:31:49.900
This is going to be my x, this is going to be my y.
00:31:49.900 --> 00:31:56.200
I will call this my a and this my b.
00:31:56.200 --> 00:32:13.900
When I turn it so that I’m looking at my triangle, I have got my triangle this way.
00:32:13.900 --> 00:32:20.900
I have got that, the axis is right through the middle.
00:32:20.900 --> 00:32:26.100
The base of the triangle, the base are equilateral triangles.
00:32:26.100 --> 00:32:33.700
This angle is 60°, I did not rotate it this time.
00:32:33.700 --> 00:32:42.400
That is the height, this is the base of the triangle.
00:32:42.400 --> 00:32:46.100
We want to find the area of the base.
00:32:46.100 --> 00:32:55.700
We know that area = base × height/ 2.
00:32:55.700 --> 00:33:03.500
If this is x and this is y, that means this right here, half of it, from here to here is y.
00:33:03.500 --> 00:33:08.500
Therefore, the base is actually equal to 2y.
00:33:08.500 --> 00:33:12.000
Let us go ahead and find y, same way as before.
00:33:12.000 --> 00:33:25.700
We have x/y, x is to y, as h is to side length is a.
00:33:25.700 --> 00:33:39.000
a/2 which implies that y is equal to x × a/ 2h.
00:33:39.000 --> 00:33:46.200
b is equal to 2y, b is equal to 2 of these.
00:33:46.200 --> 00:33:57.500
Therefore, we have just x × a/h, that is our base.
00:33:57.500 --> 00:34:03.000
Let us redraw our triangle.
00:34:03.000 --> 00:34:09.100
This was our h, this was our y, this was our b.
00:34:09.100 --> 00:34:12.600
Our h, this is a 60° angle.
00:34:12.600 --> 00:34:20.500
Therefore, if this is y, this is y√3.
00:34:20.500 --> 00:34:40.200
It = √3 × ax/ 2h.
00:34:40.200 --> 00:34:53.700
We said that the area is equal to the base × height/ 2 = the base which is ax/h
00:34:53.700 --> 00:35:12.000
× the height which is √3 × ax/ 2h.
00:35:12.000 --> 00:35:15.100
I will go ahead and put this ½ over here.
00:35:15.100 --> 00:35:30.200
Therefore, our area is √3 a² x²/ 4h².
00:35:30.200 --> 00:35:36.300
There you go, we have our area function which is a function of x.
00:35:36.300 --> 00:35:40.100
We are going to b; remember this is r, we put it this way.
00:35:40.100 --> 00:35:48.800
We are going to be integrating from 0 to h, from here to here, adding up all of our slices like that.
00:35:48.800 --> 00:36:00.600
Therefore, our volume is equal to the integral from 0 to h of our area function × dx,
00:36:00.600 --> 00:36:18.000
which is the integral from 0 to h of this thing √3 a²/ 4h² x² dx.
00:36:18.000 --> 00:36:24.200
This is what we want and the integral is easy after that.
00:36:24.200 --> 00:36:27.400
This is all a constant, it comes out of the integral.
00:36:27.400 --> 00:36:30.000
This, when you integrate it, just becomes x³/ 3.
00:36:30.000 --> 00:36:32.000
I’m going to leave the integration to you.
00:36:32.000 --> 00:36:37.600
I hope you forgive me for that, very simple integration.
00:36:37.600 --> 00:36:40.800
That is the formula that we are looking for.
00:36:40.800 --> 00:36:49.500
Let us see what else what we have got here.
00:36:49.500 --> 00:36:55.900
Let us see what we can do with this one.
00:36:55.900 --> 00:36:57.000
Let me go back to black.
00:36:57.000 --> 00:37:02.800
Find the volume of the solid whose base is given by the equation, 16x² + 4y² = 64.
00:37:02.800 --> 00:37:05.700
We are looking at a base that is an ellipse.
00:37:05.700 --> 00:37:16.400
And whose cross section is perpendicular to the y axis are isosceles right triangles,
00:37:16.400 --> 00:37:21.100
with a base of the triangle being the hypotenuse.
00:37:21.100 --> 00:37:25.700
A lot going on here, let us see what we have,
00:37:25.700 --> 00:37:29.800
whose base is given by the equation 16 x² + 4y² = 64.
00:37:29.800 --> 00:37:33.000
Let us go ahead and take care of this first.
00:37:33.000 --> 00:37:39.100
We have got 16x² + 4y² = 64.
00:37:39.100 --> 00:37:51.300
We have got x²/ 2² + y²/ 4² = 1.
00:37:51.300 --> 00:37:56.500
Let me draw it a little bit over here, in fact.
00:37:56.500 --> 00:38:02.300
I will draw it over here.
00:38:02.300 --> 00:38:09.900
Our base is this, we go 1, 2, 3, 4, 1, 2, 3, 4.
00:38:09.900 --> 00:38:13.100
We go 1, 2, we go 1, 2.
00:38:13.100 --> 00:38:22.400
We are looking at an ellipse, something like that.
00:38:22.400 --> 00:38:25.600
They say this cross section is perpendicular to the y axis.
00:38:25.600 --> 00:38:34.600
We are going to hit the y axis, in other words, we are going to hit the y axis.
00:38:34.600 --> 00:38:40.100
The cross sectional area is our isosceles right triangles.
00:38:40.100 --> 00:38:45.900
I take a cross section along the y axis.
00:38:45.900 --> 00:38:54.600
When I pull this cross section out and I flip it this way up, from your perspective, I have a cross section.
00:38:54.600 --> 00:39:00.100
I’m going to flip it this way, these are isosceles right triangles.
00:39:00.100 --> 00:39:06.800
What it is going to look like is the following.
00:39:06.800 --> 00:39:08.500
Let me actually marks some points here.
00:39:08.500 --> 00:39:15.500
If this is point a and this is point b, looking at it from the top, when I flip it up and rotate it,
00:39:15.500 --> 00:39:18.100
I’m going to get an isosceles right triangle.
00:39:18.100 --> 00:39:26.500
This is going to be an isosceles right triangle.
00:39:26.500 --> 00:39:36.700
This is a and this is b, taken the cross section, I flipped it up so I’m actually looking at it.
00:39:36.700 --> 00:39:42.200
Let us see what we have got here.
00:39:42.200 --> 00:39:46.900
Let us go ahead and call this h again.
00:39:46.900 --> 00:39:50.100
Let us go ahead and call this b.
00:39:50.100 --> 00:39:54.400
Again, we are looking at the area of the triangle.
00:39:54.400 --> 00:39:55.900
We are going to find the area of the triangle.
00:39:55.900 --> 00:39:59.900
This time, we are going to integrate along the y axis.
00:39:59.900 --> 00:40:02.500
We are going to be integrating from -4 to 4.
00:40:02.500 --> 00:40:05.800
Those are going to be our limits of integration.
00:40:05.800 --> 00:40:12.100
Again, we have the area = the base × the height/ 2.
00:40:12.100 --> 00:40:26.900
Because we are integrating along the y axis, we need a(y).
00:40:26.900 --> 00:40:32.100
We need an area function that is a function of y because we are integrating along the y axis.
00:40:32.100 --> 00:40:37.100
Let us go ahead and find a relationship.
00:40:37.100 --> 00:40:47.500
If this is our x value, this is going to be our y value, the whole idea.
00:40:47.500 --> 00:41:02.300
There is each point along the ellipse is in relationship between x and y.
00:41:02.300 --> 00:41:08.100
The x right here, that is this.
00:41:08.100 --> 00:41:11.000
This distance is our x.
00:41:11.000 --> 00:41:14.300
The base is going to be twice the x.
00:41:14.300 --> 00:41:18.000
This is the x, that is the x, that is our base.
00:41:18.000 --> 00:41:23.800
We need to find x, in terms of y.
00:41:23.800 --> 00:41:27.000
We need to multiply by 2, let us do that.
00:41:27.000 --> 00:41:34.000
16x² + 4y² = 64, I hope that made sense.
00:41:34.000 --> 00:41:38.900
The way we draw this triangle based on this thing that we have flipped up,
00:41:38.900 --> 00:41:43.600
this half the base of the triangle is our x value, whatever x is.
00:41:43.600 --> 00:41:50.800
We need to find an expression in y, we need to find the relationship between x and y.
00:41:50.800 --> 00:41:56.600
The base is twice x.
00:41:56.600 --> 00:42:06.200
We have got 16x² = 64 - 4y².
00:42:06.200 --> 00:42:14.300
We have x² = 64 – 4y²/ 16.
00:42:14.300 --> 00:42:25.600
Therefore, x = 1/4 √64 - 4y².
00:42:25.600 --> 00:42:30.900
I know that you can simplify it more but I just want to probably leave it like this, does not really matter.
00:42:30.900 --> 00:42:40.200
x is this, that is our x value and our y value is going to be this thing, whatever that happens to be.
00:42:40.200 --> 00:42:57.300
The base is equal to twice x, that is equal to ½ × √64 – 4y².
00:42:57.300 --> 00:43:01.900
We have our base.
00:43:01.900 --> 00:43:06.000
Let us see what we can do about our height.
00:43:06.000 --> 00:43:08.600
Let us draw our triangle again.
00:43:08.600 --> 00:43:13.000
We have a right isosceles triangle, there we go.
00:43:13.000 --> 00:43:25.700
We set this as the base, this is the height.
00:43:25.700 --> 00:43:37.600
If this was our x value, this is 90° that means that is 45 and that is 45.
00:43:37.600 --> 00:43:42.300
That is 45, that means this is 90.
00:43:42.300 --> 00:43:45.500
h and x are the same.
00:43:45.500 --> 00:43:54.500
h is actually equal to x and that is equal to b/2.
00:43:54.500 --> 00:44:04.900
h is equal to x, we found x that is equal to 1/4 √64 – 4y².
00:44:04.900 --> 00:44:19.400
Now we have our h, our area as a function of y is equal to ½ the base × the height = ½ of the base
00:44:19.400 --> 00:44:37.400
which we said is ½ √64 – 4y² × the height which is ¼ × √64 - 4y².
00:44:37.400 --> 00:44:41.200
Our function is equal to 2 × 2 is 4.
00:44:41.200 --> 00:44:49.600
We are going to get 1/16, 64 - 4y².
00:44:49.600 --> 00:44:52.500
You can simplify that out a little bit more, if you want.
00:44:52.500 --> 00:45:01.200
That is going to be 4 - 1/4 y².
00:45:01.200 --> 00:45:16.400
Therefore, the volume, we are going to integrate from, that was this.
00:45:16.400 --> 00:45:18.700
We are going to add up all the triangles.
00:45:18.700 --> 00:45:23.400
The area is this, we are going to go from -4 to 4.
00:45:23.400 --> 00:45:38.500
The integral from -4 to 4 of 4 - ¼ y² dy.
00:45:38.500 --> 00:45:47.800
Because this is symmetric, if you want, you can also write it as twice the integral from 0 to 4.
00:45:47.800 --> 00:45:51.000
This area is the same as that area, you can do it this way.
00:45:51.000 --> 00:45:57.100
Changing one of the limits to 0, if you want to, it is not a big deal.
00:45:57.100 --> 00:46:05.200
4 - ¼ y² dy, there you go.
00:46:05.200 --> 00:46:13.100
This is what we wanted, I hope that make sense.
00:46:13.100 --> 00:46:16.000
Let us do one more problem.
00:46:16.000 --> 00:46:24.100
Find the volume of the solid whose base is the region bounded by the function y = 3 - x² in the x axis,
00:46:24.100 --> 00:46:32.500
and whose cross section is perpendicular to the x axis are squares.
00:46:32.500 --> 00:46:36.300
Volume whose base is the region bounded by the function 3 – x² in the x axis.
00:46:36.300 --> 00:46:40.100
Let us go ahead and draw this out.
00:46:40.100 --> 00:46:43.500
This one should be reasonably straightforward.
00:46:43.500 --> 00:46:51.700
3 - x², we go 1, 2, up to 3.
00:46:51.700 --> 00:46:58.300
3 - x² in the x axis, this cross section is perpendicular to the x axis.
00:46:58.300 --> 00:47:05.300
We are going to be taking slices perpendicular to the x axis.
00:47:05.300 --> 00:47:08.500
This is the base, are squares.
00:47:08.500 --> 00:47:19.800
When I take this slice out, turn it, it is a square.
00:47:19.800 --> 00:47:31.100
s and s, these points of intersection by the way, -√3 and √3.
00:47:31.100 --> 00:47:37.300
If you are wondering where that came from, set this 3 – x² to 0.
00:47:37.300 --> 00:47:40.200
You are going to get x = + or -√3.
00:47:40.200 --> 00:47:44.600
It is going to tell me what the 0’s are, what the roots are of this equation.
00:47:44.600 --> 00:47:53.900
The area is equal to side².
00:47:53.900 --> 00:48:04.600
If this is x, if this is my value of x, this is my value of y.
00:48:04.600 --> 00:48:09.600
Therefore, that is actually equal to y.
00:48:09.600 --> 00:48:18.600
Therefore, y is equal to 3 - x², it is a function.
00:48:18.600 --> 00:48:27.900
The height of the square is 3 - x².
00:48:27.900 --> 00:48:40.600
My area which is equal to s² is equal to 3 - x²².
00:48:40.600 --> 00:48:51.900
Therefore, the area is equal to 9 - 6x² + x⁴.
00:48:51.900 --> 00:48:58.900
Our volume, we took the slice this way, we are going to be adding up all the volumes this way.
00:48:58.900 --> 00:49:16.000
It is just equal to -√3 to √3 9 – 6x² + x⁴ dx.
00:49:16.000 --> 00:49:21.000
I will leave the integration to you, whether you want to use your calculator or do it by hand.
00:49:21.000 --> 00:49:28.800
That is all, dealing with regions with a cross section is not a circle.
00:49:28.800 --> 00:49:33.400
There is no rotation, but we can still deal with it.
00:49:33.400 --> 00:49:35.500
Thank you so much for joining us here at www.educator.com.
00:49:35.500 --> 00:49:36.000
We will see you next time, bye.