WEBVTT mathematics/calculus-ii/murray
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Today we are going to learn how to find the surface area of revolution.
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What that means is that we are going to take a function y=f(x)
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We are going to revolve this function around the y axis.
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We are going to take this curve and spin it around the y axis and create a surface.
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Then, what we are asking today is if you were going to paint that surface, how much paint would it require.
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In other words, what is the surface area of the surface that you would obtain.
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We have a nice formula that tells us the answer here.
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The formula is the integral from x=a to x=b of 2π × f(x) × sqrt(1 + f'(x)²) dx.
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You might recognize part of this formula as something we saw in a previous lecture on arc length.
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Indeed, that is not a coincidence.
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The surface area formula comes from looking at a small piece of the curve,
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Calculating its arc length, and then calculating what the surface area would be if we rotated that around the x axis.
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Of course, that is where you get the 2π/f(x) part of the formula.
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Let us try this out with some examples.
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The first example we are going to be finding the surface area of the cone
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If we take the graph of y = 3x and we rotate that around the x axis from x=0 to x=2.
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We would get that cone and we are trying to figure out what the surface area is.
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Remember our formula is the integral of 2π f(x) × sqrt(1 + f'(x)²) dx.
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In this case, our f(x) is 3x.
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So f'(x) is just 3.
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1 + f'(x)² is 1 + 3² which is 10.
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So this part of the formula, the square root part, is sqrt of 10.
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We are integrating from a to b is 0 and 2, from x=0 to x=2 of 2π × f(x) is 3x.
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× the square root of 10, dx.
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That is our surface area formula.
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I am going to pull the 2π × 3 to the outside.
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That gives us 6π.
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I will pull the sqrt(10) outside as well, that is the sqrt(10).
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The integral of x is x²/2,
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And we want to evaluate that from x=0 to x=2.
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That gives us 6π × sqrt(10) × if you plug in x=2 there, you get 4 - 0.
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Sorry, 4/2 - 0.
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That is 6π × sqrt(10).
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4/2 is just 2 so 2 - 0.
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The whole answer is 12π × sqrt(10).
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Key points here, we are rotating something around the x axis.
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We look at the f(x) that we are given and we calculate this square root formula by doing f'(x) and then 1 + f'².
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We plug it all in to the surface area formula and then do the integration to finish that off.
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Let us try another example.
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This time we are rotating the graph of y = x³ around the x axis.
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Our f'(x) is 3x².
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Our f'² is 9x⁴
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1 + f'² is 1 + 9x⁴.
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And, the square root of that is sqrt(9x³ + 1).
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So, our surface area is the integral from x=0 to x=1 of 2π f(x).
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So, I put 2 π, now f(x) is x³, so 2pix³ × integral of 9x⁴ + 1 dx.
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Now, this looks like a rough integral but it is actually not so bad.
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If you notice, you have 9x⁴ + 1 under the radical.
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The derivative of that would 36x³, and we have an x³ outside.
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What we can do is let u = 9x⁴ + 1.
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Then du is 36x³ dx.
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So, what this converts into is, I guess I can write dx or x³ dx, is 1/36 du.
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That takes care of the dx and the x³
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I will put the 1/36 outside.
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I am also going to put the 2π outside.
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We still have the integral from x=0 to x=1.
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Now, 9x⁴ + 1 just turned into u, and then we have du.
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That is really a very simple integral now.
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This is now π/18, u^1/2 is the same as sqrt(u).
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If we integrate that, that integrates to u^3/2/3/2 or, 2/3u^3/2.
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We are evaluating this not using u but using x=0, not x=1.
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So let us keep going with that.
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We still have a π.
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2/3 I can cancel that with the 18 to get 9 here and a 1 there.
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π over 27.
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Now u was 9x⁴ + 1, and we are evaluating that from x=0 to x=1.
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We get π/27.
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Now if you plug in x=1, I am sorry I left out my 3/2 there.
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That 3/2 came from that right there.
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Then we have 9x⁴ + 1.
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If we plug in 1 to 9x⁴ + 1, that is 10^3/2
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X=0, if you plug it in just gives you 1 - 1^3/2.
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This gives us π/27 × 10^3/2, is the same as saying 10 sqrt(10), - 1.
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That is the answer for the surface area of this graph we rotated around the x axis.
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Again, the key point there is identifying your f(x)
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Running it through this formula, plugging it into the general surface area formula
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Then, we had a pretty tricky integral there.
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The key thing there was observing that the derivative of 9x⁴ + 1 was more or less x³,
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So we could make this substitution that gave us a very nice integral to solve.
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Let us try another example
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This time we are rotating the graph of y = 2x² + 1 from x=0 to x=1.
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There is a very important difference in this example which is we are rotating around the y axis.
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Remember that our surface area formula that we learned before used the x axis.
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We had the x axis before, that means we need to convert our surface area formula
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To adapt to the fact that we are now rotating around the y-axis.
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That means we need to look at the surface area formula and sort of switch everything from x's to y's.
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Let me write that down.
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The surface area is now the integral from y =, I will not write a and b, c to y = d, of 2π f(y).
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× sqrt(1+f'(y)) dy.
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That means we need to convert everything here into functions and terms of y.
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Including our function and the limits.
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We were given this as if y were a function of x.
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Instead we need to convert everything to x as a function of y.
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Let us go ahead and convert that.
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Looking at the function first, we get y-1 = 2x²
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We can divide both sides by 2 there, so x = sqrt(y-1/2).
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If you plug in x=0, into the function,
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That would convert into y = 1.
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If you plug x=1 again into the function,
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That would convert into y=3.
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Now we have everything in terms of y.
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Again, let us try to figure out what f'(y) and what this square root formula turns into.
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f'(y), well we have got a square root of something
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So its derivative is 1/2
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Because we think of its square root as all of that stuff to the 1/2.
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1/2 times all of that stuff to the -1/2
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I will write it down here in the denominator, y-1/2
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× the derivative of that inside stuff by the chain rule
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The derivative of y-1/2 is 1/2.
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Let me try to simplify this.
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I am going to put the two 1/2's together and put 1/4.
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Now this denominator, I am going to flip it and bring it up to the numerator
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That is 2/y-1.
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That is f'
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We need to find f'².
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f'² is 1/16 × 2/y-1.
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We can simplify that into 1/8 × y-1.
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1 + f'(y)² is, 8 × y-1, 8y-8+1/8y-1.
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What I did there is I wrote that 1 as 8y-8/the denominator.
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8y-8, because I wanted to combine everything over a common denominator.
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I can simplify that down a little bit into 8y-7/8(y-1).
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Now we need to take the square root of that.
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What we have is the square root of 1 + f'(y)².
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Is the sqrt(8y-7/8(y-1)).
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I am going to put all of these pieces together into a big integral.
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The integral is the surface area from y=1 to y=3.
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Got those from here and here.
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2π f(y), I got that from here, that is the sqrt(y-1/2) × sqrt(1+f'(y²)),
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That is 8-7/8(y-1) × dy.
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That looks pretty messy but it does simplify.
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The y-1's cancel and then we are left with a 2 and an 8 in the denominator, that is 16.
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We can pull that out of the square root and that is a 4 in the denominator,
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And so this simplifies down a little bit.
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2/4 gives us 1/2, I will pull the π outside.
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So, we get the integral from y=1 to y=3 of square root of, I think the only thing that is left there is 8y-7 dy.
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Now we can use u = 8y-7 so du = 8 dy.
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dy is 1/8 du.
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What we get now is π/2 × the 1/8 that we got from the du here.
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The integral from y=1 to y=3 of u^1/2 du.
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That is π/2 × 1/8 u^1/2
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The integral of u^1/2 is u^3/2/3/2, which is the same as multiplying by 2/3.
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And, we want to evaluate that from y=1 to y=3.
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I can simplify a little bit.
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My two's cancel, and we get π, combine the 3 and the 8 to give 24.
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That is π/24
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Now, u was 8y-7^3/2
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Finally, we get π/24 ×
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Now, if we plug in y=3 to 8y-7.
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That is 24-7, which is 17^3/2
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If we plug in y=1 to 8y-7 which is 8-7, so -1^3/2
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We can clean that up a little bit into π/24 × 17 sqrt(17).
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1^3/2 is just 1.
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Our final answer is π/24 × 17 sqrt(17) -1.
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Probably what made that problem a little difficult besides it being a little bit complicated on the algebraic side.
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Was the fact that we were given the y axis instead of the x axis
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Which means you have to take your original x formula and translate everything into terms of y.
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In turn, you have to translate things from y as a function of x to x as a function of y.
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The x values that you are given have to be converted into y values.
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Once you do that, you walk through the process of calculating this radical 1 + x'(y)².
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Then it is still kind of a messy integral but the square roots sort of cancel nicely
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And, you end up with something that is not too complicated at the end.