Thread: Finding Revolving Volumes

1. 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 about the -axis.

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.

2. 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 about the -axis.

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 $\displaystyle f(x)=e^{4x}+4$

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

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

...

5. I tried that integral. I got 30.67, and that was incorrect.

6. 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 $\displaystyle 18.17$.

7. Oh, I understand much better! I've been really struggling with this one.

Thanks. I'm so happy now!

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