I'm currently taking a Calculus III course and we're learning about triple integrals. So far I am having a lot of trouble with them. I need help figuring out how I should solve the following integral.
Calculate $\iiint_{D}\sqrt(x^2+y^2+z^2)dxdydz$ where D is the region
within the sphere $x^{2}+y^{2}+z^{2}=1$ and above the cone $z =\sqrt(x^{2}+y^{2})$.
This is how far I've come. I plugged in (x,y,z) = (0,0,0) in both functions and saw that the sphere is the "top" function. So we're integrating from the cone to the sphere. By looking at the problem I am getting the sense that I eventually want to use spherical coordinates (or cylindrical? I don't know how to judge.) to solve this problem easier but I am not entirely sure how I would go about using them. I need to get the boundaries for the outer integral first before I continue.
$x = \rho sin\phi cos\theta$
$y = \rho sin\phi sin\theta$
$z = \rho cos\phi$
$\rho^{2} = x^{2}+y^{2}+z^{2}$
$\iiint_{\sqrt(x^{2}+y^{2})}^{\sqrt(1-x^{2}-y^{2})} \sqrt(x^2+y^2+z^2) dzdydx$
This is the point where I usually get stuck with when I do triple integrals. I have no idea how I should think when trying to get the region/bounds for the outer double integral. I am struggling hard with this part with every integral I come across and I would like to learn some kind of thinking process/strategy regarding this. Any help is appreciated!
Best Answer
If you mean the solid $Q$ which is interior to the cone, you have
$$Q=\left\{(x,y,z)\,|\,\sqrt{x^2+y^2}\leq z \leq \sqrt{1-x^2-y^2}\,,\,(x,y)\in D\,\right\}$$
with $D$ the proyecction of $Q$ over $z$ plane, that is
$$D=\{(x,y)\,|\,x^2+y^2\leq\frac{1}{2}\}$$
and then your triple integral can be done (using cylindrical coordenates)
$$\iiint_{Q}\sqrt{x^2+y^2+z^2}dxdydz=\int_0^{2\pi}\int_0^{\sqrt{1/2}}\int_r^{\sqrt{1-r^2}}r\,\sqrt{r^2+z^2}\,dz\,dr\,d\theta=\frac{1}{4} \left(2-\sqrt{2}\right) \pi$$