Evaluate triple integral: int[0,2]int[0,sqrt(4-x^2)]int[....

[hank]

$\displaystyle \int^{2}_{0} \int^{\sqrt{4-x^2}}_{0} \int^{3-x^2-y^2}_{-5+x^2+y^2} x dzdydx$



I've tried the problem several times and I get to this point and I'm stuck:

$\displaystyle \int^{2}_{0} 8x \sqrt{4-x^2} -2x^3 \sqrt{4-x^2} -\frac{8}{3} x \sqrt{4-x^2} + \frac{2}{3} x^3 \sqrt{4-x^2}dx$

I'm not sure how to integrate the terms with x^3, and I have a hunch I did something wrong and those terms should drop out anyway.

 

[galactus]

Re: Evaluate the triple integral

Did you try converting to polar coordinates?. may make it a little easier.

 

[hank]

Re: Evaluate the triple integral

Hi.
I did try converting it, even though we hadn't learned out to that yet at that point.

I got zero for my answer, so I don't think that's right.

 

[tkhunny]

Re: Evaluate the triple integral

Your two lower bounds of zero (0) seem to indicate you are thinking about symmetries. Are you?

Your initial expression multiplied by 4 looks nice. What do you get?

In cyllindrical coordinates, $\displaystyle [0,\frac{\pi}{2}]$ and then multiplied by 4 looks nice. What do you get?

Watch that rogue cosine. It's periodic, you know.

 

[galactus]

Re: Evaluate the triple integral

Do the first integration and we get:

$\displaystyle \int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}\left[2x(4-x^{2})-2xy^{2}\right]dydx$

$\displaystyle =\int_{0}^{2}\frac{4}{3}x(4-x^{2})^{\frac{3}{2}}dx$

Now, finish the last one.

 

[hank]

Oh, man, I see it now.

The answer was buried in a lot of ugly algebra. Thanks!