Integrate $\iiint_E z\,dx\,dy\,dz$ in $E=\lbrace(x,y,z) : x^2+y^2+z^2 \leq 4, z \leq 1 \rbrace$

Question

I want to integrate $\iiint_E z\,dx\,dy\,dz$ in $E=\lbrace(x,y,z) : x^2+y^2+z^2 \leq 4, z \leq 1 \rbrace$.

I want to split that triple integral in 3 integrals, but I don't see which the limits are.

I guess I have to do $x \leq \pm \sqrt{4-y^2-z^2}$ and integrate in $x$ between those limits (I don't know if this is correct) but then I get lost with the $y$ limits. If I let $z$ be the last variable I integrate, I suppose I can say that $z$ is between $-2$ and $1$, so that limits are easy.

What are the limits of the $y$ integral? And are the $x$ limits okay? Thanks in advance.

Answer 1

I will first tell you the easiest way which is using cylindrical coordinates.

$x^2+y^2 = r^2$ so $x^2+y^2+z^2 \leq 4 \implies r^2 \leq 4-z^2$

So the integral becomes,

$\displaystyle \int_0^{2\pi} \int_{-2}^{1} \int_0^{\sqrt{4-z^2}} r \ z \ dr \ dz \ d\theta$

Can you see why it is the simplest?

Also note that as $z$ is an odd function, its integral between $-1 \leq z \leq 1$ will be simply zero due to symmetry of the region wrt z-plane.

So integral between $-2 \leq z \leq -1$ will give you the same result. This point becomes important if you decide to use spherical coordinates for the integral.

If one is integrating for $z \leq 1$ using spherical coordinates, the integral must be split into two. In spherical coordinates,

$x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$

At $z = 1, \rho = \sec \phi$ and to find upper bound of $\phi$, $1 = 2 \cos\phi \implies \phi = \frac{\pi}{3}$. So the integral becomes,

$\displaystyle \int_0^{2\pi} \int_0^{\pi/3} \int_{0}^{\sec\phi} \rho^3 \cos\phi \sin\phi \ d\rho \ d\phi \ d\theta + \int_0^{2\pi} \int_{\pi/3}^{\pi} \int_{0}^{2} \rho^3 \cos\phi \sin\phi \ d\rho \ d\phi \ d\theta$

Now as I mentioned earlier, you can simply write it as one integral between $-2 \leq z \leq -1$ and I will leave the set-up of that integral as an exercise for you.

Answer 2

I suggest you use spherical coordinates in which a Cartesian triple $(x,y,z)$ is equivalent to a spherical triple $(R,\theta,\phi)$ such that $${ x=R\sin\theta\cos\phi\\ x=R\sin\theta\sin\phi\\ z=R\cos\theta.} $$ In this spherical coordinates, you have $$ {x^2+y^2+z^2\le 4\iff R\le 2 \\ z\le 1 \iff R\cos\theta\le 1 \\ dxdydz=R^2\sin\theta d\theta d\phi dR. } $$ Now substitute and proceed to integration.

Answer 3

I would split the integral into two integrals, one of length and the other of area: Since $z$ varies in the interval $[-2,1]$, the first integral should be $$\int^1_{-2}(\cdots)dz$$then if you fix $z$, you have $x^2+y^2\leqslant 4-z^2$, so you have a closed disc $B_z$ on the plane of center $(0,0)$ and radius $\sqrt{4-z^2}$, so the integral inside should be $$\iint_{B_z}zdxdy$$ so your integral is equal to $$\int^1_{-2}\iint_{B_z}zdxdydz=*$$

Now you can take $z$ out of the double integral, so $$*=\int^1_{-2}z\left(\iint_{B_z}dxdy\right)dz=\int^1_{-2}z\pi(4-z^2)dz$$ (since $\iint_{B_z}dxdy$ is noting but the area of the disc $B_z$, that is, $\pi r^2=\pi(\sqrt{4-z^2})^2=\pi(4-z^2)$)