Chapter 6.2, Problem 25E

a.

To determine

Find the volume enclosed by the given integral $\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{2}xdx}$

Answer to Problem 25E

The volume enclosed by the given integral is $\left\{\left(x,y\right)0\le x\le \frac{\pi }{2},0\le y\le \cos \left(x\right)\right\}$

Explanation of Solution

Calculation:

Consider the equation $\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{2}xdx}$

AS the given integral represents the volume of the solid, we know that it is:

  $\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\left(\cos \left(x\right)\right)}^{2}dx}={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)}dx$

If the cross- section is a disk, we know that $A\left(x\right)=\pi {\left(\text{radius}\right)}^2$

If $x$ is read from the integral bounds and $y$ is the radius of the cross- section

Then the region is: $\left\{\left(x,y\right)0\le x\le \frac{\pi }{2},0\le y\le \cos \left(x\right)\right\}$

Thus, the Find the volume enclosed by the given integral $\left\{\left(x,y\right)0\le x\le \frac{\pi }{2},0\le y\le \cos \left(x\right)\right\}$

b.

To determine

Find the volume enclosed by the given integral $\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(y^4-y^8\right)dy}$

Answer to Problem 25E

The volume enclosed by the given integral $\left\{\left(x,y\right)0\le y\le 1,y^4\le x\le y^2\right\}.$

Explanation of Solution

Calculation:

Consider the equation $\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(y^4-y^8\right)dy}$

AS the given integral represents the volume of the solid, we know that it is:

  $\pi {\displaystyle \underset{0}{\overset{1}{\int }}{\left(y^4-y^8\right)}^{2}dy}=\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left({\left(y^2\right)}^2-{\left(y^4\right)}^2\right)}dx={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)}dx.$

If the cross- section is a washer, we know that $A\left(x\right)=\pi {\left(\text{outer radius}\right)}^2-\pi {\left(\text{inner radius}\right)}^2$.

If $y$ is read from the integral bounds and $x$ is the radius of the cross- section

Then the region is: $\left\{\left(x,y\right)0\le y\le 1,y^4\le x\le y^2\right\}.$

Thus, the volume enclosed by the given integral $\left\{\left(x,y\right)0\le y\le 1,y^4\le x\le y^2\right\}.$