# Thread: Application of Integration Problem

1. ## [Solved] Application of Integration Problem

A ceramic bowl is turned on a potter's wheel to approximate the surface obtained by revolving the curve y= (2/3)((x^2) - 1 ) from x=1 to x=2 about the y axis. What is the approximate outside area of the bowl? [assume that each unit on the x and y axes represents 10 cm]

Any thoughts?

I'd really appreciate any help, especially if it would lead to the solution.

Thanks.

2. ## Re: Application of Integration Problem

Originally Posted by andvaka
A ceramic bowl is turned on a potter's wheel to approximate the surface obtained by revolving the curve y= (2/3)((x^2) - 1 ) from x=1 to x=2 about the y axis. What is the approximate outside area of the bowl? [assume that each unit on the x and y axes represents 10 cm]

Any thoughts?
We are finding the area of a surface of revolution:

$S = 2\pi\int_a^br(x)\sqrt{1+\left(\frac{dy}{dx}\right) ^2}\,dx$

$= 2\pi\int_1^2x\sqrt{1+\left(\frac43x\right)^2}\,dx$

3. ## Re: Application of Integration Problem

Originally Posted by Reckoner
We are finding the area of a surface of revolution:

$S = 2\pi\int_a^br(x)\sqrt{1+\left(\frac{dy}{dx}\right) ^2}\,dx$

$= 2\pi\int_1^2x\sqrt{1+\left(\frac43x\right)^2}\,dx$

But shouldn't r(x) be y instead of x? Correct me if I am wrong. I don't understand how you got x from r(x).

$=2\pi\int_{1}^{2}(\frac{2}{3}(x^2 -1))\sqrt{1+(\frac{4x}{3})^2} dx$

4. ## Re: Application of Integration Problem

Originally Posted by andvaka
But shouldn't r(x) be y instead of x? Correct me if I am wrong. I don't understand how you got x from r(x).

$=2\pi\int_{1}^{2}(\frac{2}{3}(x^2 -1))\sqrt{1+(\frac{4x}{3})^2} dx$
$r(x)$ represents the distance between the curve and the axis of revolution at a given $x$-value. Since we are revolving the curve about the $y$-axis, that distance is just $x.$ If we were revolving about the $x$-axis, then you would be correct: the distance from the curve to the axis would be $y = \frac23\left(x^2-1\right).$

So if we were revolving about the $x$-axis, the area of the surface would be

$S = 2\pi\int_1^2\frac23\left(x^2-1\right)\sqrt{1+\left(\frac{4x}3\right)^2}\,dx.$

Consider another example, let's say we wanted to revolve the curve about the line x = 3. Then the distance from the curve to the axis of revolution would be $r(x) = 3 - x$ and the area would be

$S = 2\pi\int_1^2(3 - x)\sqrt{1+\left(\frac{4x}3\right)^2}\,dx.$

Give me a second and I'll make a picture to help you visualize it.

5. ## Re: Application of Integration Problem

Ah. Understood. I got confused because of the notation used in my textbook vs yours.
Thanks!

6. ## Re: Application of Integration Problem

Please click the thumbnail below for a larger version.

The graph shows the curve along with the three different axes of revolution that I mentioned in my last post. The distance between the point $(x, y)$ and each axis of revolution is also depicted. The problem that you posted involved revolution about the line $x = 0$ (that is, the $y$-axis), which is shown in red.

Sorry for the notational confusion. I should have explained the formula I was using more clearly.