Chapter 6.2, Problem 45E

a.

To determine

Find the volume enclosed by the given integral.

Answer to Problem 45E

The volume enclosed by the given integral $8\pi R{\displaystyle \underset{0}{\overset{r}{\int }}\sqrt{r^2-y^2}dy}$

Explanation of Solution

Consider the equation $13x^2+6\sqrt{3}xy+7y^2=16$

Calculation:

  

The volume can be found by rotating a circle around the $y-$ axis.the

Equation of the a circle of radius $r$ centered at $\left(R,0\right)$ is ${\left(x-R\right)}^2+y^2=r^2$

To rotate around the $y$ -axis using the washer method, solve for $x$

  $\begin{array}{l}{\left(x-R\right)}^2=r^2-y^2\\ x-R=\pm \sqrt{r^2-y^2}\\ x=R\pm \sqrt{r^2-y^2}\end{array}$

The $+$ one is the right half; the $-$ one is the left of the circle.

We can use symmetry and double the integration along the $y$ -axis from

  $0\text{ to }r$ . Outer radius will be $r_1=R+\sqrt{r^2-y^2}$

Inner radius is $r_2=R-\sqrt{r^2-y^2}$

Subtract inner from outer

  $\begin{array}{l}V=2{\displaystyle \underset{a}{\overset{b}{\int }}\pi \left(r_1^2-r_2^2\right)}dy\\ =2\pi {\displaystyle \underset{0}{\overset{r}{\int }}\left({\left(R+\sqrt{r^2-y^2}\right)}^2-{\left(R-\sqrt{r^2-y^2}\right)}^2\right)}dy\\ =2\pi {\displaystyle \underset{0}{\overset{r}{\int }}\left(R^2+2R\sqrt{r^2-y^2}+r^2-y^2\right)-\left(R^2-2R\sqrt{r^2-y^2}+r^2-y^2\right)dy}\\ =2\pi {\displaystyle \underset{0}{\overset{r}{\int }}\left(R^2+2R\sqrt{r^2-y^2}+r^2-y^2-R^2+2R\sqrt{r^2-y^2}-r^2+y^2\right)}dy\\ =2\pi {\displaystyle \underset{0}{\overset{r}{\int }}4R}\sqrt{r^2-y^2}dy\\ =8\pi R{\displaystyle \underset{0}{\overset{r}{\int }}\sqrt{r^2-y^2}dy}\end{array}$

Thus, the volume enclosed by the given integral $8\pi R{\displaystyle \underset{0}{\overset{r}{\int }}\sqrt{r^2-y^2}dy}$

b.

To determine

Find the volume enclosed by the given integral.

Answer to Problem 45E

The volume enclosed by the given integral $2{\pi }^2{r}^{2}R$

Explanation of Solution

From part (a)

  $V=8\pi R{\displaystyle \underset{0}{\overset{r}{\int }}\sqrt{r^2-y^2}dy}$

If $x=\sqrt{r^2-y^2}$ then square both sides,

  $\begin{array}{l}{x}^2=r^2-y^2\\ x^2+y^2=r^2\end{array}$ Equation of the circle

So $x=\sqrt{r^2-y^2}$ represents the right half.

We can replace the integral with the area of a circle $\pi r^2$ ,multiply by $\frac{1}{4}$ .

  $\begin{array}{l}V=8\pi R{\displaystyle \underset{0}{\overset{r}{\int }}\sqrt{r^2-y^2}dy}\\ =8\pi R\left(\frac{1}{4}\cdot \pi r^2\right)\\ =2{\pi }^2{r}^{2}R\end{array}$

Thus, the volume enclosed by the given integral $2{\pi }^2{r}^{2}R$