Need Help: Volume of Ellipse Rotated at X=-2 Using the Shell Method

[Pull and Twist]

The question is...

Find the volume of the solid obtained by rotating the ellipse \(\displaystyle {x}^{2}+4{y}^{2}=4\) about \(\displaystyle x=-2\) using the Shell Method. *Table of Integrals is not allowed for integration.

So of course I rearrange my ellipse formula to get \(\displaystyle y=\pm\sqrt{1-\frac{{x}^{2}}{4}}\)

Then I calculate my radius as \(\displaystyle x-(-2)=r \implies x+2=r\)

I know the formula for finding volume with the Shell Method is \(\displaystyle V=\int_{a}^{b}2\pi x\cdot f(x) \,dx\)

To make it easier I decide to just find the volume of the top half and times my equation by two. Having graphed the equation I know I need to integrate between \(\displaystyle -2\le x\le2\)

So I then have...

\(\displaystyle 2\int_{-2}^{2} 2\pi(x+2)(\sqrt{1-\frac{{x}^{2}}{4}})\,dx\)

\(\displaystyle 4\pi\left\{\int_{-2}^{2} x(\sqrt{1-\frac{{x}^{2}}{4}})\,dx + \int_{-2}^{2} 2(\sqrt{1-\frac{{x}^{2}}{4}})\,dx\right\}\)

\(\displaystyle 4\pi\left\{\int_{-2}^{2} x(\sqrt{1-\frac{{x}^{2}}{4}})\,dx + 2\int_{-2}^{2}(\sqrt{1-\frac{{x}^{2}}{4}})\,dx\right\}\)

Using \(\displaystyle u=1-\frac{{x}^{4}}{4}\) substitution for the first integral I get...

\(\displaystyle 4\pi\left\{-\frac{4}{3}\left[{\left(1-\frac{{x}^{2}}{4}\right)}^{\frac{3}{2}}\right]_{-2}^{2} + 2\int_{-2}^{2}(\sqrt{1-\frac{{x}^{2}}{4}})\,dx\right\}\)

Am I taking the right approach?? If so... how do I integrate the second integral w/o using a table of integrals?? Would this problem be easier utilizing the disk/washer method?

I need me some help. Thanks.

 

[MarkFL]

First off, we know this is a torus, and so the volume will be:

\(\displaystyle V=2\pi(2)\pi(2)(1)=8\pi^2\)

We can use this to check our answer. So to set this problem up, I would begin with the volume of an arbitrary shell:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=x+2\)

\(\displaystyle h=2y=2\frac{\sqrt{4-x^2}}{2}=\sqrt{4-x^2}\)

And so we find:

\(\displaystyle V=2\pi\int_{-2}^2(x+2)\sqrt{4-x^2}\,dx=2\pi\left(\int_{-2}^2 x\sqrt{4-x^2}\,dx+2\int_{-2}^2\sqrt{4-x^2}\,dx\right)\)

For the first integral, I would use the odd-function rule (it is zero), and for the second integral I would use the even function rule to get:

\(\displaystyle V=8\pi\int_{0}^2\sqrt{4-x^2}\,dx\)

Next, I would use the substitution:

\(\displaystyle x=2\sin(\theta)\,\therefore\,dx=2\cos(\theta)\)

Or you could argue that the remaining integral represents the area of a quarter circle having radius 2. :D

Can you proceed?

 

[Pull and Twist]

Awesome, that worked! I didn't even think about substituting \(\displaystyle x=2\sin\left({\theta}\right)\). I've been spoiled by using my Table of Integrals.

That left me with \(\displaystyle {\cos}^{2}\left({\theta}\right)\) after utilizing my trig identities and combining terms.

Replaced that using the half angle formula... substituted \(\displaystyle u=2\theta \) \(\displaystyle \frac{du}{2}=dx\) and integrated the result.

\(\displaystyle 8\pi\left[u+\sin\left({u}\right)\right]_{0}^{\pi}\)

Which gives you \(\displaystyle 8\pi\left[\pi\right]\) or \(\displaystyle 8{\pi}^{2}\)