Question

Evaluate ∫∫∫z2dx dydz
S
,where S is the solid region between the spheres ρ =1 and
ρ = 2, by using spherical coordinates.

Answer 1

Answer on Question #70103 – Math – Real Analysis

Question

Evaluate $\iiint_{S} z^{2} dx dy dz$ where $S$ is the solid region between the spheres $\rho = 1$ and $\rho = 2$, by using spherical coordinates.

Solution

Let us use the spherical coordinates

(see http://mathworld.wolfram.com/SphericalCoordinates.html):

$$\left\{ \begin{array}{l} x = r \cos \theta \sin \varphi \\ y = r \sin \theta \sin \varphi, \; r \in [1, 2], \; \theta \in [0, 2\pi), \; \varphi \in [0, \pi]. \end{array} \right.$$

The Jacobian is $\left| \frac{\partial(x,y,z)}{\partial(r,\theta,\varphi)} \right| = r^2 \sin \varphi$. Then

the original integral is equal to $\iiint_{D} r^{4} (\cos \varphi)^{2} \sin \varphi \, dr \, d\theta \, d\varphi =$

$$\begin{array}{l} \int_{1}^{2} \int_{0}^{\pi} \int_{0}^{2\pi} r^{4} (\cos \varphi)^{2} \sin \varphi \, d\theta \, d\varphi \, dr = \left( \int_{1}^{2} r^{4} dr \right) \left( \int_{0}^{\pi} (\cos \varphi)^{2} \sin \varphi \, d\varphi \right) \left( \int_{0}^{2\pi} d\theta \right) = \\ = 2\pi \left( \frac{r^{5}}{5} \Big|_{r=1}^{2} \right) \left( -\int_{0}^{\pi} (\cos \varphi)^{2} d(\cos \varphi) \right) = -\frac{62\pi}{5} \left( \frac{(\cos \varphi)^{3}}{3} \Big|_{0=0}^{\pi} \right) = \frac{124\pi}{15} = 8\frac{4}{15} \pi. \end{array}$$

Answer: $8\frac{4}{15}\pi$.

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