Chapter 6.2, Problem 44E

a.

To determine

Find the volume enclosed by the given integral.

Answer to Problem 44E

The volume enclosed by the given integral $2{\displaystyle \underset{0}{\overset{r}{\int }}h\sqrt{r^2-x^2}}dx$

Explanation of Solution

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

Calculation:

  

Circle with radius $r$ centered at the origin; $x^2+y^2=r^2$

Solved for y: $y=\pm \sqrt{r^2-x^2}$

The top half is the positive square root . which gives the length of half

The base of one isosceles triangle cross-section ( $AD$ in the example cross section

In the diagram above) .multiply by $2$ to get the whole base length $AB$ base $=2\sqrt{r^2-x^2}$

The height $h$ is the same for every triangle so the area of one cross-section is

  $A\left(x\right)=\frac{1}{2}\cdot 2\sqrt{r^2-x^2}\cdot h=h\sqrt{r^2-x^2}$

The circle intersect the $x-$ at $x=-r\text{ and }x=r$ so we could integrate with those limits. it is symmetric though so we can integrate from $0\text{ to }r$ ,

And multiply by $2$ instead $V=2{\displaystyle \underset{0}{\overset{r}{\int }}h\sqrt{r^2-x^2}}dx$

Thus, the volume enclosed by the given integral $2{\displaystyle \underset{0}{\overset{r}{\int }}h\sqrt{r^2-x^2}}dx$

b.

To determine

Find the volume enclosed by the given integral.

Answer to Problem 44E

The volume enclosed by the given integral $\frac{1}{2}\pi r^{2}h$

Explanation of Solution

From part (a)

  $\begin{array}{l}V=2h{\displaystyle \underset{0}{\overset{r}{\int }}\sqrt{r^2-x^2}}dx\\ \end{array}$

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

  $\begin{array}{l}{y}^2=r^2-x^2\\ r^2=x^2+y^2\text{ equation of a circle}\end{array}$

So $y=\sqrt{r^2-x^2}$ represents the upper half of a circle of radius $r$ centered

At the origin .the integral limits $\text{0 to r}$ then is half of the of circle ( $1/4$ of a whole circle).

We can replace the integral with the area of a circle $\pi r^2,$ multiplied by $1/4$ .

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

Thus, the volume enclosed by the given integral is $\frac{1}{2}\pi r^{2}h$