Thread: Describing/sketching region of integration of triple integral

1. Describing/sketching region of integration of triple integral

Hi there,

I'm having great difficulty with the following problem:

This question concerns the integral $\int_{0}^{2}\int_{0}^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}\!z\ \mathrm{d}z\ \mathrm{d}x\ \mathrm{d}y$. Sketch or describe in words the domain of integration. Rewrite the integral in both cylindrical and spherical coordinates. Which is easier to evaluate?
Below is what I believe I have established so far...

The projection of this integral's domain onto the $xy$-plane is the portion of the circle $x^2+y^2=4$ on $0\le x\le2,\ y\ge0$.

The bounds on $z$ correspond to

$z^2=x^2+y^2$ (cone) and $x^2+y^2+z^2=8$ (sphere).
These bounds intersect at

$x^2+y^2=4$.
Below $z=2$ (where the bounds on $z$ intersect), I believe that the cone and cylinder, $x^2+y^2=4$, are completely inside the sphere.

Would it hence be correct to say that the region of integration is the solid lying between the cone and the cylinder, on $x\ge0$, $y\ge0$ and $0\le z\le2$? I'm struggling to visualize this problem.

When I attempt to move on, and evaluate the integral in cylindrical/spherical coordinates, my solutions differ by a factor of 2.

That is, I evaluated this integral as,

$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{0}^{ \sqrt{8-r^2}}\!z\ r\ \mathrm{d}z\ \mathrm{d}r\ \mathrm{d}\theta=2\pi$
And,

$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi=4\pi$

Thank you very much.

2. Re: Describing/sketching region of integration of triple integral

Hey drokkin.

Can you fix up your tex code? I can't really make out what the equations are.

3. Re: Describing/sketching region of integration of triple integral

Originally Posted by chiro
Hey drokkin.

Can you fix up your tex code? I can't really make out what the equations are.
Sorry - hit the "Submit" button instead of the "Preview" button! All sorted now.