Question

a) Find the volume inside the cone$\mathbf{3}{\mathit{z}}^{\mathbf{2}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{y}$, above the plane$\mathit{z}\mathbf{=}\mathbf{2}$ and inside the sphere ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathbf{36}$. Hint: Use spherical coordinates.

b) Find the centroid of the volume in (a)

Short answer

(a). The required volume is $64\tau \tau$.

(b). The centroid is $\left(0,0,\frac{231}{64}\right)$.

Step-by-step solution

Given information

The solid is bounded by the cone $3z^2=x^2+y$, the plane $z=2$and the sphere $x^2+y^2+z^2=36$.

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:

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

Substitute 0 for   in the equation of cone.

(a)

Calculate the volume of cone as follows:

$$\begin{gathered} z^2=\frac{x^2}{3}z \\ =\frac{x^2}{\sqrt{3}}\alpha \\ =\arctan \left(\frac{x}{z}\right) \\ =\frac{\tau \tau }{3} \end{gathered}$$

The required volume is as follows:

$$\begin{gathered} V={\int }_{\frac{2}{\cos \theta }}^6{r}^{2}dr{\int }_0^{\tau \tau /3}\sin \theta d\theta {\int }_0^{2\tau \tau }d\varphi \\ =\frac{2\tau \tau }{3}{\left[r^3\right]}_{\frac{2}{\cos \theta }}^0{\int }_0^{\tau \tau /3}\sin \theta {\left[r^3\right]}_{\frac{2}{\cos \theta }}^{6}d\theta \\ =\frac{2\tau \tau }{3}{\left(108-\frac{4}{{\cos }^2\theta }\right)}_0^{\tau \tau /3} \\ =64\tau \tau \end{gathered}$$

Thus, the required volume is$64\tau \tau$ .

Use rcosθ  for z in the formula for centroid

(b)

Here, $$\begin{gathered} \overline{x}=\overline{y}=0 \\ \overline{z}=\frac{1}{V}{\int }_{v}zdV \\ =\frac{1}{64\tau \tau }{\int }_{\frac{2}{\cos \theta }}{r}^{3}dr{\int }_0^{\tau \tau /3}\sin \theta \cos \theta d\theta {\int }_0^{2\tau \tau }d\varphi \\ =\frac{1}{128}{\left[-324\cos \left(2\theta \right)-\frac{8}{{\cos }^2\theta }\right]}_0^{\tau \tau /3} \\ =\frac{231}{64} \end{gathered}$$.

Thus, the centroid is $\left(0,0,\frac{231}{64}\right)$.