Question

Write a triple integral in cylindrical coordinates for the volume inside the cylinder ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{4}$and between $\mathit{z}\mathbf{=}\mathbf{2}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}$ and the (x,y) plane. Evaluate the integral.

Short answer

The required volume is$12\tau \tau$

Step-by-step solution

Given information

The solid is bounded by the cylinder the cylinder $x^2+y^2=4$ and between $z=2x^2+y^2$and the (x,y) plane.

Concept of spherical coordinates

The spherical coordinates $\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{,}\mathbf{\varphi }\mathbf{)}$ is related to Cartesian coordinates (x,y,z) by:

$$\begin{gathered} \mathit{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{\theta }} \\ \mathbf{}\mathbf{}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{\right)}\mathit{\varphi } \\ \mathbf{}\mathbf{}\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{z}}{\mathbf{r}}\mathbf{\right)} \end{gathered}$$

The cylindrical coordinates$\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{,}\mathbf{\varphi }\mathbf{)}$ is related to Cartesian coordinates(x,y,z) by:

$$\begin{gathered} \mathit{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{\theta }} \\ \mathbf{}\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{\right)}\mathit{z} \\ \mathbf{}\mathbf{=}\mathit{z} \end{gathered}$$

The limits of polar coordinates

The required integral is I ${\int }_0^{2}dx{\int }_0^{\sqrt{4-x^2}}dy{\int }_0^{2x^2+y^2}dz$.

$$\begin{gathered} r:0\to 2 \\ \theta :0\to 2\tau \tau \\ z:0\to 2r^2{\cos }^2\theta +r^2\left({\cos }^2\theta +1\right) \end{gathered}$$

Calculate further as follows:

$$\begin{gathered} I={\int }_0^{2}dx{\int }_0^{\sqrt{4-x^2}}dy{\int }_0^{2x^2+y^2}dz \\ ={\int }_0^{2}rdr{\int }_0^{2\tau \tau }d\theta {\int }_0^{r^2\left({\cos }^2\theta +1\right)}dz \\ ={\int }_0^{2}rdr{\int }_0^{2\tau \tau }{r}^2\left({\cos }^2\theta +1\right)d\theta \end{gathered}$$

The above integral can be written as follows:

$$\begin{gathered} I={\int }_0^2{r}^{3}dr{\int }_0^{2\tau \tau }{r}^2\left({\cos }^2\theta +1\right)d\theta \\ ={\left(\frac{r^4}{4}\right)}_0^2\left(2\tau \tau +\frac{1}{2}{\int }_0^2\left(\cos 2\theta +1\right)d\theta \right) \\ =4\left(2\tau \tau +\tau \tau \right) \\ =12\tau \tau \end{gathered}$$

Therefore, the required volume is $12\tau \tau$.