Describing/sketching region of integration of triple integral

[drokkin]

Hi there,

I'm having great difficulty with the following problem:

This question concerns the integral $\displaystyle \int_{0}^{2}\int_{0}^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}\!z\ \mathrm{d}z\ \mathrm{d}x\ \mathrm{d}y$. 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 $\displaystyle xy$-plane is the portion of the circle $\displaystyle x^2+y^2=4$ on $\displaystyle 0\le x\le2,\ y\ge0$.

The bounds on $\displaystyle z$ correspond to

$\displaystyle z^2=x^2+y^2$ (cone) and $\displaystyle x^2+y^2+z^2=8$ (sphere).

These bounds intersect at

$\displaystyle x^2+y^2=4$.

Below $\displaystyle z=2$ (where the bounds on $\displaystyle z$ intersect), I believe that the cone and cylinder, $\displaystyle x^2+y^2=4$, 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 $\displaystyle x\ge0$, $\displaystyle y\ge0$ and $\displaystyle 0\le z\le2$? 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,

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

And,

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

Can you please help me to identify where I am going wrong?

Thank you very much.

 

[daon2]

$\displaystyle z$ is trapped between the cone and the sphere. You will get the shape of an "ice-cream cone" where the circle of intersection as you described is at z=2. Excuse the bad drawing

 

[drokkin]

$\displaystyle z$ is trapped between the cone and the sphere. You will get the shape of an "ice-cream cone" where the circle of intersection as you described is at z=2. Excuse the bad drawing

View attachment 2901

I'm still struggling to identify what steps you've taken to establish the region that you've sketched. It feels like I'm missing a fairly simple concept here... Do you think you can offer any further explanation?

From the region that you've sketched, I take it that the bounds that I have put on $\displaystyle \phi$ are incorrect and should range from $\displaystyle 0$ to $\displaystyle \frac{\pi}{4}$. The fact that $\displaystyle \int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi=2\pi$ seems to confirm this.

Thank you very much for your help!

 

[daon2]

I'm still struggling to identify what steps you've taken to establish the region that you've sketched. It feels like I'm missing a fairly simple concept here... Do you think you can offer any further explanation?

From the region that you've sketched, I take it that the bounds that I have put on $\displaystyle \phi$ are incorrect and should range from $\displaystyle 0$ to $\displaystyle \frac{\pi}{4}$. The fact that $\displaystyle \int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi=2\pi$ seems to confirm this.

Thank you very much for your help!

The integral is telling you that $\displaystyle \sqrt{x^2+y^2} < z < \sqrt{8-x^2-y^2}$ translating directly to "$\displaystyle z$ lies above the surface of the cone and below the surface of the sphere". The "cylinder" you speak of is not part of the bounded solid, rather, its intersection with the xy-plane gives a bound for x and y.