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Professor Murray

Professor Murray

Surface Area of Revolution

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Table of Contents

I. Advanced Integration Techniques
Integration by Parts

24m 52s

Intro
0:00
Important Equation
0:07
Where It Comes From (Product Rule)
0:20
Why Use It?
0:35
Lecture Example 1
1:24
Lecture Example 2
3:30
Shortcut: Tabular Integration
7:34
Example
7:52
Lecture Example 3
10:00
Mnemonic: LIATE
14:44
Ln, Inverse, Algebra, Trigonometry, e
15:38
Additional Example 4
-1
Additional Example 5
-2
Integration of Trigonometric Functions

25m 30s

Intro
0:00
Important Equation
0:07
Powers (Odd and Even)
0:19
What To Do
1:03
Lecture Example 1
1:37
Lecture Example 2
3:12
Half-Angle Formulas
6:16
Both Powers Even
6:31
Lecture Example 3
7:06
Lecture Example 4
10:59
Additional Example 5
-1
Additional Example 6
-2
Trigonometric Substitutions

30m 9s

Intro
0:00
Important Equations
0:06
How They Work
0:35
Example
1:45
Remember: du and dx
2:50
Lecture Example 1
3:43
Lecture Example 2
10:01
Lecture Example 3
12:04
Additional Example 4
-1
Additional Example 5
-2
Partial Fractions

41m 22s

Intro
0:00
Overview
0:07
Why Use It?
0:18
Lecture Example 1
1:21
Lecture Example 2
6:52
Lecture Example 3
13:28
Additional Example 4
-1
Additional Example 5
-2
Integration Tables

20m

Intro
0:00
Using Tables
0:09
Match Exactly
0:32
Lecture Example 1
1:16
Lecture Example 2
5:28
Lecture Example 3
8:51
Additional Example 4
-1
Additional Example 5
-2
Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule

22m 36s

Intro
0:00
Trapezoidal Rule
0:13
Graphical Representation
0:20
How They Work
1:08
Formula
1:47
Why a Trapezoid?
2:53
Lecture Example 1
5:10
Midpoint Rule
8:23
Why Midpoints?
8:56
Formula
9:37
Lecture Example 2
11:22
Left/Right Endpoint Rule
13:54
Left Endpoint
14:08
Right Endpoint
14:39
Lecture Example 3
15:32
Additional Example 4
-1
Additional Example 5
-2
Simpson's Rule

21m 8s

Intro
0:00
Important Equation
0:03
Estimating Area
0:28
Difference from Previous Methods
0:50
General Principle
1:09
Lecture Example 1
3:49
Lecture Example 2
6:32
Lecture Example 3
9:07
Additional Example 4
-1
Additional Example 5
-2
Improper Integration

44m 18s

Intro
0:00
Horizontal and Vertical Asymptotes
0:04
Example: Horizontal
0:16
Formal Notation
0:37
Example: Vertical
1:58
Formal Notation
2:29
Lecture Example 1
5:01
Lecture Example 2
7:41
Lecture Example 3
11:32
Lecture Example 4
15:49
Formulas to Remember
18:26
Improper Integrals
18:36
Lecture Example 5
21:34
Lecture Example 6 (Hidden Discontinuities)
26:51
Additional Example 7
-1
Additional Example 8
-2
II. Applications of Integrals, part 2
Arclength

23m 20s

Intro
0:00
Important Equation
0:04
Why It Works
0:49
Common Mistake
1:21
Lecture Example 1
2:14
Lecture Example 2
6:26
Lecture Example 3
10:49
Additional Example 4
-1
Additional Example 5
-2
Surface Area of Revolution

28m 53s

Intro
0:00
Important Equation
0:05
Surface Area
0:38
Relation to Arclength
1:11
Lecture Example 1
1:46
Lecture Example 2
4:29
Lecture Example 3
9:34
Additional Example 4
-1
Additional Example 5
-2
Hydrostatic Pressure

24m 37s

Intro
0:00
Important Equation
0:09
Main Idea
0:12
Different Forces
0:45
Weight Density Constant
1:10
Variables (Depth and Width)
2:21
Lecture Example 1
3:28
Additional Example 2
-1
Additional Example 3
-2
Center of Mass

25m 39s

Intro
0:00
Important Equation
0:07
Main Idea
0:25
Centroid
1:00
Area
1:28
Lecture Example 1
1:44
Lecture Example 2
6:13
Lecture Example 3
10:04
Additional Example 4
-1
Additional Example 5
-2
III. Parametric Functions
Parametric Curves

22m 26s

Intro
0:00
Important Equations
0:05
Slope of Tangent Line
0:30
Arc length
1:03
Lecture Example 1
1:40
Lecture Example 2
4:23
Lecture Example 3
8:38
Additional Example 4
-1
Additional Example 5
-2
Polar Coordinates

30m 59s

Intro
0:00
Important Equations
0:05
Polar Coordinates in Calculus
0:42
Area
0:58
Arc length
1:41
Lecture Example 1
2:14
Lecture Example 2
4:12
Lecture Example 3
10:06
Additional Example 4
-1
Additional Example 5
-2
IV. Sequences and Series
Sequences

31m 13s

Intro
0:00
Definition and Theorem
0:05
Monotonically Increasing
0:25
Monotonically Decreasing
0:40
Monotonic
0:48
Bounded
1:00
Theorem
1:11
Lecture Example 1
1:31
Lecture Example 2
11:06
Lecture Example 3
14:03
Additional Example 4
-1
Additional Example 5
-2
Series

31m 46s

Intro
0:00
Important Definitions
0:05
Sigma Notation
0:13
Sequence of Partial Sums
0:30
Converging to a Limit
1:49
Diverging to Infinite
2:20
Geometric Series
2:40
Common Ratio
2:47
Sum of a Geometric Series
3:09
Test for Divergence
5:11
Not for Convergence
6:06
Lecture Example 1
8:32
Lecture Example 2
10:25
Lecture Example 3
16:26
Additional Example 4
-1
Additional Example 5
-2
Integral Test

23m 26s

Intro
0:00
Important Theorem and Definition
0:05
Three Conditions
0:25
Converging and Diverging
0:51
P-Series
1:11
Lecture Example 1
2:19
Lecture Example 2
5:08
Lecture Example 3
6:38
Additional Example 4
-1
Additional Example 5
-2
Comparison Test

22m 44s

Intro
0:00
Important Tests
0:01
Comparison Test
0:22
Limit Comparison Test
1:05
Lecture Example 1
1:44
Lecture Example 2
3:52
Lecture Example 3
6:01
Lecture Example 4
10:04
Additional Example 5
-1
Additional Example 6
-2
Alternating Series

25m 26s

Intro
0:00
Main Theorems
0:05
Alternation Series Test (Leibniz)
0:11
How It Works
0:26
Two Conditions
0:46
Never Use for Divergence
1:12
Estimates of Sums
1:50
Lecture Example 1
3:19
Lecture Example 2
4:46
Lecture Example 3
6:28
Additional Example 4
-1
Additional Example 5
-2
Ratio Test and Root Test

33m 27s

Intro
0:00
Theorems and Definitions
0:06
Two Common Questions
0:17
Absolutely Convergent
0:45
Conditionally Convergent
1:18
Divergent
1:51
Missing Case
2:02
Ratio Test
3:07
Root Test
4:45
Lecture Example 1
5:46
Lecture Example 2
9:23
Lecture Example 3
13:13
Additional Example 4
-1
Additional Example 5
-2
Power Series

38m 36s

Intro
0:00
Main Definitions and Pattern
0:07
What Is The Point
0:22
Radius of Convergence Pattern
0:45
Interval of Convergence
2:42
Lecture Example 1
3:24
Lecture Example 2
10:55
Lecture Example 3
14:44
Additional Example 4
-1
Additional Example 5
-2
V. Taylor and Maclaurin Series
Taylor Series and Maclaurin Series

30m 18s

Intro
0:00
Taylor and Maclaurin Series
0:08
Taylor Series
0:12
Maclaurin Series
0:59
Taylor Polynomial
1:20
Lecture Example 1
2:35
Lecture Example 2
6:51
Lecture Example 3
11:38
Lecture Example 4
17:29
Additional Example 5
-1
Additional Example 6
-2
Taylor Polynomial Applications

50m 50s

Intro
0:00
Main Formulas
0:06
Alternating Series Error Bound
0:28
Taylor's Remainder Theorem
1:18
Lecture Example 1
3:09
Lecture Example 2
9:08
Lecture Example 3
17:35
Additional Example 4
-1
Additional Example 5
-2
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Lecture Comments (8)

1 answer

Last reply by: Dr. William Murray
Fri Apr 15, 2016 5:20 PM

Post by Gautham Padmakumar on April 13, 2016

At 10:43, I noticed that you made a mistake when you wrote down the integral formula about the y axis. You left it as f'(y) instead of (f'(y))^2

Thanks for the lecture series by the way! It really helped me review for my calc finals

1 answer

Last reply by: Dr. William Murray
Tue Feb 11, 2014 4:18 PM

Post by Katrina Forrest on February 8, 2014

Professor Murray,

I just want to comment on the brilliancy of your lectures. You present the material in a way that is structured, concise, and comprehendible. I wish more of my instructors followed your style of teaching!

1 answer

Last reply by: Dr. William Murray
Fri Sep 6, 2013 12:35 PM

Post by A De Lama on September 2, 2013

WOnderful lectures, by the way. This really is the future of education!!

1 answer

Last reply by: Dr. William Murray
Fri Sep 6, 2013 12:34 PM

Post by Anubis De Lamali on September 2, 2013

when solving for x in terms of y, why is it not "+ or -" in front of the square root?
THanks!!

Surface Area of Revolution

Main formula:

Surface Area of Revolution =

Hints and tips:

  • To remember this formula, it helps to recall that the square root part comes from the arclength formula (which in turn comes from the Pythagorean Theorem). The 2πf (x) part comes from the circumference of a circle, 2πr, where the radius r is the height of the curve f (x) that is revolving around the x-axis.

  • Remember that you must integrate the square root formula above. A common mistake is to integrate the function itself, not the square root formula. Of course, this would give you the area under the curve and not the surface area of revolution.

  • A similar mistake is to mix this up with the arclength formula, which looks similar. Be careful which one you are asked for.

  • Don’t make the common algebraic mistake of thinking that reduces to a + b! This is extremely wrong, and your teacher will likely be merciless if you do it

  • A common technique in problems of this nature is to make a u-substitution for whatever is under the square root sign. Then (hopefully) you can manipulate the expression outside the square root (which comes from f (x)dx) into being the du. However, you might have to do several steps of algebraic manipulation, pulling factors in or out of the square root sign, before this works.

  • If the graph is being revolved around the y-axis, simply switch the roles of x and y in the formula above. Be sure to check carefully which one the problem is asking for.

  • When it’s feasible, check that your answer makes sense. Unlike area integrals, which can be negative if a curve goes below the x-axis, surface area of revolution should always be positive!

Surface Area of Revolution

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Important Equation 0:05
    • Surface Area
    • Relation to Arclength
  • Lecture Example 1 1:46
  • Lecture Example 2 4:29
  • Lecture Example 3 9:34
  • Additional Example 4
  • Additional Example 5

Transcription: Surface Area of Revolution

I am going to do a couple more examples of surface area of revolution.0000

We are going to start out by rotating the graph of 2x3 from x=0 to x=1 around the x axis.0004

So here, our f(x) is 2x3, we want to find f'(x), which is 6x2,0012

f'(x)2 is 36 x4.0025

1 + that is 36x4 + 1.0034

The square root of 1 + f'(x)2 is just the square root of 36x4 + 1.0041

The point of doing that was that you want to use your surface area formula.0054

That involves this big radical expression.0057

Our surface area is equal to the integral from x=0 to x=1, of 2pi × f(x), which is 2x3.0060

× the radical expression, 36x4 + 1 × dx.0079

Now, I am going to pull the 4pi outside.0092

x=0 to x=1.0097

Key thing to notice here, this 36x4 looks pretty nasty, but its derivative is exactly x3.0101

Well not exactly x3, but x3 times a constant.0110

That suggests the substitution, u = 36x4 + 1.0115

Then du is 4 × 36 x3 dx.0123

We kind of have the du here when we have x3 dx.0131

The only thing that is not quite right is the 4 × 36.0134

We can correct for that by dividing on the outside by 4 × 36.0139

Then we will have the x3 dx gives you du.0145

We now have the square root of 36x3 + 1, that is u.0149

So this is pi/36.0156

Integral of sqrt(u) is not too bad, you think of that as u1/2.0161

So the integral is u3/2 and divide by 3/2, which is the same as multiplying by 2/3.0169

We want to evaluate this from x=0 to x=1.0180

We cannot do that directly because everything is still in terms of u.0184

I am going to write this as pi, I guess we can simplify the 2 and 36 to be 1 and 18.0190

Then 18 × 3 is 54.0198

Now u3/2, (36x4 + 1)3/2, and we want to evaluate that from x=0 to x=1.0202

That is pi/54, now if we plug in x=1 to 36x4 + 1, we will get 37 3/2.0218

- x=0 in there just gives you 13/2.0235

This can be slightly simplified to pi/54 ×, 373/2 is the same as saying 37×sqrt(37)0243

- 1/23/2 is just 1.0253

So we get our final answer there.0258

So, the key part of that problem is identifying y=f(x).0266

Then working through the formula, finding f'(x) and figuring out what the square root of 1 +f'2 is.0270

Then plugging the whole thing into the surface area formula that we learned at the beginning of the lecture.0280

Then it looks like a tricky integral but the key thing is to notice that if we let u = 36 x4 + 1,0286

We basically have our du set up for us with 2x3, we just need to correct for the constant there.0294

For our last example, I would like to find the surface area by rotating the graph of y=sqrt(2x) around the x axis.0000

Here, our f(x) is sqrt(2x), and f'(x),0009

If you think of f(x) is 2x1/2, then f'(x) is 1/2 × 2x-1/2 × the derivative of 2x,0015

So those 2's cancel and you get 1/sqrt(2x).0031

Now f'(x)2 is just 1/2x, and if we add 1 to that, we get,0038

If we put those over a common denominator we get 2x+1/2x.0053

Finally the sqrt(1+f'(x)2) is sqrt(2x+1/2x).0061

Now we are ready to invoke our surface area of revolution formula.0070

The surface area of revolution is the integral on the bounds we are given are x=0 to x=1, of 2pi × f(x),0075

Which is the sqrt(2x).0088

× this big square root formula, 2x+1/2x.0091

All integrated with respect to dx.0100

So, this is actually pretty nice because the sqrt(2x) here, cancels with sqrt(2x) in the denominator there.0106

Now we have the integral of sqrt(2x+1), and a natural thing to do there is to let u=2x+1.0115

Then a du is just 2dx.0130

I am going to pull a 2pi outside, we get 2pi × integral from x=0 to x=1.0135

Of the sqrt(u) and dx converts into du, except that dx is 1/2 du.0145

I will put that 1/2 outside.0154

Sorry the integral from x=0 to x=1, of sqrt(u) du.0157

So, the point of that was that sqrt(u) is an easy integral.0172

You think of that as u1/2, and the integral of u1/2 is u3/2/3/20180

Which is the same as multiplying by 2/3.0193

And we are going to evaluate that from x=0 to x=1.0196

But you cannot evaluate that yet because we have the thing in terms of u, and we need to convert back into x's.0198

We remember that u=2x+1, so we get 2pi/3, 2x+13/2, evaluated from x=0 to x=1.0206

So this is 2pi/3, now if you plug in x=1 to 2x+1, that turns into two times 1+1, so 33/2,0223

- x=0 in there, into 2x+1, just gives you 13/2.0240

Again it simplifies down to 2pi/3.0247

33/2 is the same as 3sqrt(3) - 13/2 is just 1.0250

And we get our final answer.0262

To recap there, we identified f(x), that was y=sqrt(2x),0265

We plugged that into the formula to sqrt(1+f'(x)2),0270

Then we plugged our answer into the whole surface area formula right here.0277

Luckily that simplified down a little bit, then to get the integral we just needed to substitute u = 2x+1.0283

That finishes off that example and our lecture on surface area formulas is finished.0292

Today we are going to learn how to find the surface area of revolution.0000

What that means is that we are going to take a function y=f(x)0006

We are going to revolve this function around the y axis.0016

We are going to take this curve and spin it around the y axis and create a surface.0022

Then, what we are asking today is if you were going to paint that surface, how much paint would it require.0039

In other words, what is the surface area of the surface that you would obtain.0045

We have a nice formula that tells us the answer here.0052

The formula is the integral from x=a to x=b of 2pi × f(x) × sqrt(1 + f'(x)2) dx.0054

You might recognize part of this formula as something we saw in a previous lecture on arc length.0072

Indeed, that is not a coincidence.0078

The surface area formula comes from looking at a small piece of the curve,0080

Calculating its arc length, and then calculating what the surface area would be if we rotated that around the x axis.0085

Of course, that is where you get the 2pi/f(x) part of the formula.0097

Let us try this out with some examples.0104

The first example we are going to be finding the surface area of the cone0107

If we take the graph of y = 3x and we rotate that around the x axis from x=0 to x=2.0112

We would get that cone and we are trying to figure out what the surface area is.0128

Remember our formula is the integral of 2pi f(x) × sqrt(1 + f'(x)2) dx.0134

In this case, our f(x) is 3x.0147

So f'(x) is just 3.0150

1 + f'(x)2 is 1 + 32 which is 10.0156

So this part of the formula, the square root part, is sqrt of 10.0162

We are integrating from a to b is 0 and 2, from x=0 to x=2 of 2pi × f(x) is 3x.0165

× the square root of 10, dx.0178

That is our surface area formula.0183

I am going to pull the 2pi × 3 to the outside.0188

That gives us 6pi.0192

I will pull the sqrt(10) outside as well, that is the sqrt(10).0197

The integral of x is x2/2,0203

And we want to evaluate that from x=0 to x=2.0205

That gives us 6pi × sqrt(10) × if you plug in x=2 there, you get 4 - 0.0212

Sorry, 4/2 - 0.0228

That is 6pi × sqrt(10).0232

4/2 is just 2 so 2 - 0.0236

The whole answer is 12pi × sqrt(10).0239

Key points here, we are rotating something around the x axis.0247

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'2.0253

We plug it all in to the surface area formula and then do the integration to finish that off.0262

Let us try another example.0270

This time we are rotating the graph of y = x3 around the x axis.0272

Our f'(x) is 3x2.0279

Our f'2 is 9x40288

1 + f'2 is 1 + 9x4.0297

And, the square root of that is sqrt(9x3 + 1).0304

So, our surface area is the integral from x=0 to x=1 of 2pi f(x).0313

So, I put 2 pi, now f(x) is x3, so 2pix3 × integral of 9x4 + 1 dx.0326

Now, this looks like a rough integral but it is actually not so bad.0342

If you notice, you have 9x4 + 1 under the radical.0348

The derivative of that would 36x3, and we have an x3 outside.0352

What we can do is let u = 9x4 + 1.0358

Then du is 36x3 dx.0368

So, what this converts into is, I guess I can write dx or x3 dx, is 1/36 du.0376

That takes care of the dx and the x30390

I will put the 1/36 outside.0394

I am also going to put the 2pi outside.0398

We still have the integral from x=0 to x=1.0403

Now, 9x4 + 1 just turned into u, and then we have du.0408

That is really a very simple integral now.0416

This is now pi/18, u1/2 is the same as sqrt(u).0420

If we integrate that, that integrates to u3/2/3/2 or, 2/3u3/2.0429

We are evaluating this not using u but using x=0, not x=1.0442

So let us keep going with that.0450

We still have a pi.0452

2/3 I can cancel that with the 18 to get 9 here and a 1 there.0455

Pi over 27.0461

Now u was 9x4 + 1, and we are evaluating that from x=0 to x=1.0468

We get pi/27.0480

Now if you plug in x=1, I am sorry I left out my 3/2 there.0485

That 3/2 came from that right there.0490

Then we have 9x4 + 1.0495

If we plug in 1 to 9x4 + 1, that is 103/20500

X=0, if you plug it in just gives you 1 - 13/2.0507

This gives us pi/27 × 103/2, is the same as saying 10 sqrt(10), - 1.0516

That is the answer for the surface area of this graph we rotated around the x axis.0535

Again, the key point there is identifying your f(x)0543

Running it through this formula, plugging it into the general surface area formula0548

Then, we had a pretty tricky integral there.0554

The key thing there was observing that the derivative of 9x4 + 1 was more or less x3,0556

So we could make this substitution that gave us a very nice integral to solve.0568

Let us try another example0575

This time we are rotating the graph of y = 2x2 + 1 from x=0 to x=1.0578

There is a very important difference in this example which is we are rotating around the y axis.0584

Remember that our surface area formula that we learned before used the x axis.0591

We had the x axis before, that means we need to convert our surface area formula0605

To adapt to the fact that we are now rotating around the y-axis.0611

That means we need to look at the surface area formula and sort of switch everything from x's to y's.0615

Let me write that down.0621

The surface area is now the integral from y =, I will not write a and b, c to y = d, of 2pi f(y).0625

× sqrt(1+f'(y)) dy.0640

That means we need to convert everything here into functions and terms of y.0648

Including our function and the limits.0654

We were given this as if y were a function of x.0659

Instead we need to convert everything to x as a function of y.0663

Let us go ahead and convert that.0667

Looking at the function first, we get y-1 = 2x20669

We can divide both sides by 2 there, so x = sqrt(y-1/2).0679

If you plug in x=0, into the function,0688

That would convert into y = 1.0693

If you plug x=1 again into the function,0700

That would convert into y=3.0704

Now we have everything in terms of y.0710

Again, let us try to figure out what f'(y) and what this square root formula turns into.0713

f'(y), well we have got a square root of something0722

So its derivative is 1/20728

Because we think of its square root as all of that stuff to the 1/2.0732

1/2 times all of that stuff to the -1/20736

I will write it down here in the denominator, y-1/20739

× the derivative of that inside stuff by the chain rule0745

The derivative of y-1/2 is 1/2.0750

Let me try to simplify this.0754

I am going to put the two 1/2's together and put 1/4.0757

Now this denominator, I am going to flip it and bring it up to the numerator0760

That is 2/y-1.0765

That is f'0770

We need to find f'2.0772

f'2 is 1/16 × 2/y-1.0776

We can simplify that into 1/8 × y-1.0782

1 + f'(y)2 is, 8 × y-1, 8y-8+1/8y-1.0790

What I did there is I wrote that 1 as 8y-8/the denominator.0809

8y-8, because I wanted to combine everything over a common denominator.0818

I can simplify that down a little bit into 8y-7/8(y-1).0826

Now we need to take the square root of that.0837

What we have is the square root of 1 + f'(y)2.0840

Is the sqrt(8y-7/8(y-1)).0853

I am going to put all of these pieces together into a big integral.0863

The integral is the surface area from y=1 to y=3.0867

Got those from here and here.0874

2pi f(y), I got that from here, that is the sqrt(y-1/2) × sqrt(1+f'(y2)),0878

That is 8-7/8(y-1) × dy.0894

That looks pretty messy but it does simplify.0905

The y-1's cancel and then we are left with a 2 and an 8 in the denominator, that is 16.0906

We can pull that out of the square root and that is a 4 in the denominator,0913

And so this simplifies down a little bit.0919

2/4 gives us 1/2, I will pull the pi outside.0922

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.0930

Now we can use u = 8y-7 so du = 8 dy.0945

dy is 1/8 du.0955

What we get now is pi/2 × the 1/8 that we got from the du here.0962

The integral from y=1 to y=3 of u1/2 du.0974

That is pi/2 × 1/8 u1/20986

The integral of u1/2 is u3/2/3/2, which is the same as multiplying by 2/3.0993

And, we want to evaluate that from y=1 to y=3.1004

I can simplify a little bit.1008

My two's cancel, and we get pi, combine the 3 and the 8 to give 24.1010

That is pi/241019

Now, u was 8y-73/21022

Finally, we get pi/24 ×1034

Now, if we plug in y=3 to 8y-7.1044

That is 24-7, which is 173/21047

If we plug in y=1 to 8y-7 which is 8-7, so -13/21055

We can clean that up a little bit into pi/24 × 17 sqrt(17).1063

13/2 is just 1.1075

Our final answer is pi/24 × 17 sqrt(17) -1.1080

Probably what made that problem a little difficult besides it being a little bit complicated on the algebraic side.1085

Was the fact that we were given the y axis instead of the x axis1093

Which means you have to take your original x formula and translate everything into terms of y.1098

In turn, you have to translate things from y as a function of x to x as a function of y.1106

The x values that you are given have to be converted into y values.1112

Once you do that, you walk through the process of calculating this radical 1 + x'(y)2.1122

Then it is still kind of a messy integral but the square roots sort of cancel nicely1133

And, you end up with something that is not too complicated at the end.1136

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