I have to evaluate the following integral by changing to cylindrical coordinates. Should I just substitute all the $x,y,z$ to cylindrical coordinates?
$\int_{-2}^2\int_{-\sqrt{4-y^2}}^\sqrt{4-y^2}\int_\sqrt{x^2+y^2}^2 xzdzdxdy$
$\endgroup$ 12 Answers
$\begingroup$The region $(x,y,z) :-2 \leq y \leq 2, -\sqrt{4-y^2} \leq x \leq \sqrt{4-y^2}, \sqrt{x^2 + y^2} \leq z \leq 2$ is the region above the cone $ z = \sqrt{x^2 + y^2}$ and below the plane $z = 2$. In polar coordinates, this region is $(r, \theta, z): 0 \leq r \leq 2, 0 \leq \theta \leq 2\pi, r \leq z \leq 2$. If $I$ is the value of the integral,
$$\begin{align} I &= \int_0^{2\pi} \int_0^2 \int_r^2 z \cdot \underbrace{r \cos \theta}_{x} \,dz \,r \, dr \,d\theta \\ &= \left[ \int_0^{2\pi} \cos \theta \,d\theta \right] \cdot \left[ \int_0^2 \int_r^2 r^2 z \,dz \,dr \right] \\ &= \left[ \sin \theta \right]_0^{2\pi} \cdot \left[ \int_0^2 \int_r^2 r^2 z \,dz \,dr \right] \\ &= 0 \end{align}$$
$\endgroup$ $\begingroup$Let's first evaluate the integral as is:
$$\begin{align}I &= \int_{-2}^2 dy \, \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dx \, x \, \int_{\sqrt{x^2+y^2}}^2 dz \, z\\ &= \int_{-2}^2 dy \, \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dx \, x \,\left (2 - x^2-y^2 \right )\\ &= 0 \end{align} $$
because we are integrating an odd function over a symmetric interval. Now let's try this in polar coordinates:
$$\begin{align}I &= \underbrace{\int_{0}^{2 \pi} d\theta \, \cos{\theta}}_{\text{=0}} \, \int_{0}^{2} dr \, r^2 \, \int_{r}^2 dz \, z = 0 \end{align} $$
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