How to calculate the hemisphere using triple integrals?

Question

How to solve this $\displaystyle\iiint _K (y + x^2) \mathrm{d}x\mathrm{d}y\mathrm{d}z$
where $K$ is the hemisphere $x^2 + y^2 + z^2 \leq 4$, $z \geq 0$.

This is what I tried so far:

I used variable substitution

$$\iiint_K (r \sin(\phi)\sin(\theta) + r^2 \sin^2(\phi)\cos^2(\theta)) r^2 \sin(\phi) \mathrm{d}r\mathrm{d}\phi\mathrm{d}\theta$$

Here is $K$ given by the new variables:

$ 0 \leq r \leq 2$ and $ 0 \leq \theta \leq 2\pi$ and $0 \leq \phi \leq \pi$

The problem is I have no idea how to integrate this.

Answer 1

In general, instead of just naively substituting and following the nose, it is a good strategy to draw a picture and see what's happening. If the integrand is not spherically symmetric, then there is not much use of spherical polar coordinates as it might lead to uglier calculations.

Instead , first notice that , you have

$$\iint_{D}\int_{0}^{\sqrt{4-x^{2}-y^{2}}}y+x^{2}\,dz \,dxdy$$

Where $D$ is the disc $x^{2}+y^{2}\leq 4$

So you have $\iint_{D}(y+x^{2})\sqrt{4-x^{2}-y^{2}}\,dxdy$

Now you can take Polar coordinates

$$\int_{0}^{2\pi}\int_{0}^{2}(r\sin(\theta)+r^{2}\cos^{2}(\theta))\sqrt{4-r^{2}}\,r\,\,dr\,d\theta$$

Now obviously integral $\sin(\theta)$ is $0$ . So you can freely ignore the first term

All you have to do now is $\int_{0}^{2\pi}\int_{0}^{2}\cos^{2}(\theta)r\cdot r^{2}\sqrt{4-r^{2}}\,dr\,d\theta$

Which is much easier isn't it ? Substitute $r^{2}=z$ to see that

$$\frac{\pi}{2}\int_{0}^{4} z\sqrt{4-z}\,dz$$ which now is basically a one variable calculus problem which can be evaluated using integration by parts or if you know the beta function, then you can use that $

That is $$16\pi\int_{0}^{1}t\sqrt{1-t}\,dt=\frac{16\cdot 4\pi}{15}$$

So what you might have learnt from the above is that we could have considered Cylindrical Polar Coordinates to make our life easier.

Answer 2

As pointed in @A.Goodier's comments, the integral to calculate is rather $$I:=\iiint_{0\le r\le 2,0\le\theta\le2\pi,\atop0\le\phi\le\color{red}{\frac\pi2}}\left(r \sin\phi\sin\theta+ r^2 \sin^2\phi\cos^2\theta\right)r^2 \sin\phi\, \mathrm{d}r\mathrm{d}\phi\mathrm{d}\theta.$$ Here is the follow up of our dialog in comment:

$I=I_1+I_2$ where $I_1=0$ (because $\int_0^{2\pi}\sin\theta\,\mathrm d\theta=0$), so $$\begin{align}I=I_2&=\int_0^2r^4\,\mathrm dr \int_0^{2\pi}\cos^2\theta\,\mathrm d\theta\int_0^{\pi/2}\sin^3\phi\,\mathrm d\phi\\ &=\frac{2^5}5\int_0^{2\pi}\frac{1+\cos(2\theta)}2\,\mathrm d\theta \int_0^{\pi/2}(1-\cos^2\phi)\sin\phi\,\mathrm d\phi\\ &=\frac{2^5}5\pi\int_0^1(1-t^2)\,\mathrm dt \\ &=\frac{2^5}5\pi\frac23\\&=\frac{64\pi}{15}. \end{align}$$