# Finding Revolving Volumes

• Nov 9th 2009, 08:07 PM
C.C.
Finding Revolving Volumes
I've been able to calculate a lot of my homework involving finding the volume of a revolving function around either the x or y axis. However, this one seemed to pose a challenge for me.

Find the volume of the solid formed by rotating the region enclosed by

If anyone could provide an explanation and solution to this problem, it would be greatly appreciated. I'd really like to understand what this problem is asking for.

Thanks!

-C.C.
• Nov 9th 2009, 09:35 PM
redsoxfan325
Quote:

Originally Posted by C.C.
I've been able to calculate a lot of my homework involving finding the volume of a revolving function around either the x or y axis. However, this one seemed to pose a challenge for me.

Find the volume of the solid formed by rotating the region enclosed by

If anyone could provide an explanation and solution to this problem, it would be greatly appreciated. I'd really like to understand what this problem is asking for.

Thanks!

-C.C.

This is a standard problem using the shell method:

$\displaystyle V=2\pi\int_a^b xf(x)\,dx$

Plug in $\displaystyle a=0$, $\displaystyle b=0.7$ (the bounds) and $\displaystyle f(x)=e^{4x}+4$
• Nov 10th 2009, 06:40 PM
C.C.
I think I'm having problems integrating:

Integral of x(e^(4x)dx

I used integration by parts, and I got:

x((1/4)e^(4x)) minus the integral of (1/4)e^(4x)+4x dx

And I found the antiderivative of the integral. I got:

(1/16)e^(4x)+2x^2

What did I do wrong? :(
• Nov 10th 2009, 06:53 PM
redsoxfan325
Quote:

Originally Posted by C.C.
I think I'm having problems integrating:

Integral of x(e^(4x)dx

I used integration by parts, and I got:

x((1/4)e^(4x)) minus the integral of (1/4)e^(4x)+4x dx

And I found the antiderivative of the integral. I got:

(1/16)e^(4x)+2x^2

What did I do wrong? :(

$\displaystyle u=x$, $\displaystyle dv=e^{4x}\,dx$
$\displaystyle du=dx$, $\displaystyle v=\frac{1}{4}e^{4x}$

$\displaystyle \frac{xe^{4x}}{4}-\frac{1}{4}\int e^{4x}\,dx$

...
• Nov 10th 2009, 07:09 PM
C.C.
I tried that integral. I got 30.67, and that was incorrect.
• Nov 10th 2009, 07:19 PM
redsoxfan325
Quote:

Originally Posted by C.C.
I tried that integral. I got 30.67, and that was incorrect.

Sorry I should have been clearer.

$\displaystyle \int x(e^{4x}+4)\,dx=\int xe^{4x}+\int 4x\,dx$

I was only showing you how to integrate the first integral (on the right side).

Taking what I had from before, we get that the whole thing is

$\displaystyle \frac{xe^{4x}}{4}-\frac{e^{4x}}{16}+2x^2\bigg|_0^{0.7}$

Plug in the bounds and you get $\displaystyle 2.89$. Multiply by $\displaystyle 2\pi$ to get the final answer of $\displaystyle 18.17$.
• Nov 10th 2009, 07:26 PM
C.C.
Oh, I understand much better! I've been really struggling with this one.

Thanks. :) I'm so happy now!