Question

I'm trying to figure out how to evaluate a triple integral, posted below in a second. The first integration step is easy, but the second, I'm not so sure about.

Answer 1

\[\int\limits_{0}^{2}\int\limits_{0}^{4-x^{2}}\int\limits_{0}^{x} \frac {\sin(2z)}{4-z}dydzdx\]

Answer 2

Integrating first with respect to y, you get ysin(2z)/(4-z); integrating w.r.t. z is where I get caught here. Any ideas/hints?

Answer 3

Whoops, and evaluating y, it just turns into x. Should've gone further. But that still doesn't help me on the second and third integral.

Answer 4

\[\int\limits_{0}^{2}\int\limits_{0}^{4-x^{2}} \frac {x\sin(2z)}{4-z} dzdx\]

Answer 5

change it to dxdz

Answer 6

Oh, okay, derp. Didn't even think about changing order of integration. One moment.

Answer 7

\[\int\limits_{0}^{2}\int\limits_{0}^{4x^{2}} \frac {xsin(2z)}{4 - z} = \int\limits_{0}^{2}\int\limits_{0}^{\sqrt{z/4}}\]

Answer 8

(Same integrand other than switching the variables of itnegration, but I'm worried about that top left, outermost integral's upper bound. I'm not sure how to find it, it's obviously a constant and I can intuitively *think* it's two, but I'm not so sure about proving that. Anyways, I'll run with that for now unless corrected and integrate.)

Answer 9

\[\frac {x^{2}\sin(2z)}{2(4 - z)} = \frac {zsin(2z)}{8(4-z)}\]

Answer 10

you need to sketch the bounds first

Answer 11

sketch the region bounded by below curves :
z = 4-x^2
z = 0
x = 0
x = 2