1. ## Triple Integral Help

I have a couple integrals I am having trouble setting up. I would greatly appreciate any help setting these up, I'll be able to solve them no problem.
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1.) "Find the volume of the solid in the first octant bounded by x+y+z=1 and x+y+2z=1."

2.) "Evaluate the triple integral of f(x,y,z) = z(x2+y2+z2)-3/2 over the part of the ball x2+y2+z2 < 16 with z > 2."

Thanks again for looking!

2. ## Re: Triple Integral Help

Do you know how to integrate by parts? If not here's a khan academy vid explaining how it's done https://www.khanacademy.org/math/cal...-parts-formula

I don't know how useful this will be but I find this technique is useful when you have to integrate the product of 2 or more different things like (x^2)*ln(x)

EDIT: I just realized that your problem has 3 variables, and I'm not used to solving stuff like that...

3. ## Re: Triple Integral Help

I am familiar with integration by parts. Thank you for looking.

4. ## Re: Triple Integral Help

x+y+z= 1 and x+ y+ 2z= 1 intersect on the line x+y= 1 in the xy- plane (z= 0). On course, for any (x, y), the plane x+y+z= 1 is "higher" (z larger) than the plane x+y+ 2z= 1.

Projecting into the xy-plane, we have the triangle with sides the two axes, x= 0 and y= 0, and the line x+ y= 1 in the xy-plane. The line x+ y= 1 crosses the x-axis (y= 0) at (1, 0) so we can take x from 0 to 1. For each x, y goes from 0 to 1- x. And, finally, for every (x, y), z goes from (1- x- y)/2 to 1- x- y. that is the integral is
$\int_{z= (1- x- y)/2}^{1- x- y}\int_{y= 0}^{1- x}\int_{z= (1- x- y)/2}^{1- x- y} dzdydx$

For the second problem, seeing the " $x^2+ y^2+ z^2$" in the integrand, as well as the sphere $x^2+ y^2+ z^2= 16$, I would be inclined to put it into spherical coordinates. In spherical coordinates, $z= \rho cos(\phi)$ so that the plane z=2 is given by $cos(\phi)= \frac{2}{\rho}$. Of course, the sphere $x^2+ y^2+ z^2= 16$ is $\rho= 4$. Thus, integrating over that set, $0\le \phi\ cos^{-1}\left(\frac{2}{\rho}\right)$, $0\le \theta\le 2\pi$, $0\le \rho\le 4$, and the integral is $\int_{\rho= 0}^4\int_{\theta= 0}^2\pi\int_{\phi=0}^{cos^{-1}\left(\frac{2}{\rho}\right)} cos(\phi)\rho^{-1/2} d\phi d\theta d\rho$

5. ## Re: Triple Integral Help

Thank you very much!