# Thread: Volume - revolving solid

1. ## Volume - revolving solid

I am having difficulty with this question. Would someone be able to detail the steps involved in doing this question. If someone could also provide an intuitive tutorial source related to this question that would be great (book i'm using isn't so useful).

Find the volume of the solid obtained by revolving y = 4 - x^2 about the
x-axis from x = 0 to x = 2.

Edit: Does anyone know of an online calculator for these types of problems?

2. Originally Posted by anon_404
I am having difficulty with this question. Would someone be able to detail the steps involved in doing this question. If someone could also provide an intuitive tutorial source related to this question that would be great (book i'm using isn't so useful).

Find the volume of the solid obtained by revolving y = 4 - x2 about the
x-axis from x = 0 to x = 2.

Edit: Does anyone know of an online calculator for these types of problems?
I doubt that you would be allowed an "online calculator" on a test so go easy on using one until you know how to do these problems yourself!

First, draw a graph of y= 4- x^2 (NOT "x2", please) and imagine rotating that around the x- axis. You should be able to see that any point on the graph goes in a circle around the x-axis and the radius of that circle is the y coordinate of the point: radius r= y= 1- x^2. That circle has area $\pi$ r^2= $\pi$(1- x^2). You can think of that as a very thin disk with thickness " $\Delta x$" and so volume $\pi$(1- x^2) $\Delta x$. You find the volume of the entire thing by adding up all of those disks: $\pi\Sigma$(1- x^2) $\Delta x$. That is what you have learned as a "Riemann sum" and, in the limit, becomes the integral $\pi \int_0^2$(1- x^2) dx.

Can you do that integral?

3. How can i get rid of the 1 - part of the integrand?

Edit: y = 4 - x^2

4. I think halls of ivy accidentally replaced the 4 with a 1.

Why would you want to get rid of the 1? its part of the integrand, and needs to be integrated.

5. I was trying to get $x^2$ as the integrand so i could then apply an integral rule. I must be on the wrong track, how would this be integrated?

6. Wouldn't it be $r^2= \pi(4 - x^2)^2$ if $y = r = 4 - x^2$

7. Yes, you are right. Thanks for the correction. The correct integral is
$\pi\int_0^2 (4- x^2)^2 dx= \pi\int_0^2 (16- 8x^2+ x^4)dx$

And that is just
$16\pi\int_0^2 dx- 8\pi\int_0^2 x^2dxf+ \pi\int_0^2 x^4 dx$

8. Would the answer be $32\pi\frac{64\pi}{3}+\frac{32\pi}{5}$

There is suppose to be a minus after $32\pi$