Chapter 7, Problem 22RE

i.

To determine

To calculate: To solve a volume of the bowl generated by revolving the region.

Answer to Problem 22RE

  $V=4\pi$ .

Explanation of Solution

Given information: The given information as follows:

a volume of the solid generated by revolving the region enclosed by the parabola $y=\frac{x^2}{2}$ and the from $y=0toy=2$ To find the volume of the bowl.

Calculation:

The volume of the bowl generated by revolving the region bounded by the given parabola and the straight line can be calculated as:

Here, $y=\frac{x^2}{2},y=0,y=2$ .

The volume can be calculated as follows:

  $\begin{array}{c}V=\pi {\displaystyle {\int }_0^2\left(x^2\right)}dy\\ V=\pi {{\displaystyle {\int }_0^2\left(\sqrt{2y}\right)}}^{2}dy\\ V=\pi {\displaystyle {\int }_0^2\left(2y\right)}dy\\ V=2\pi {\left[\frac{y^2}{2}\right]}_0^2\\ V=\pi \left[4-0\right]\\ V=4\pi \end{array}$

ii.

To determine

To calculate: To solve a volume of the bowl generated by revolving the region.

Answer to Problem 22RE

  $V=\pi k^2$ .

Explanation of Solution

Given information: The given information as follows:

a volume of the solid generated by revolving the region enclosed by the parabola $y=\frac{x^2}{2}$ and the from $y=0toy=2$ The bowl is holding when it is filled to a depth of $k$ units.

Calculation:

The volume of the solid generated by revolving the region bounded by the given parabola and the straight line can be calculated as:

  $V={\displaystyle {\int }_a^b\pi \left(x^2\right)}dy$

Here, $y=\frac{x^2}{2},y=0,y=2$ .

When it is filled to a depth of $k$ units.

The volume can be calculated as follows:

  $\begin{array}{c}V=\pi {\displaystyle {\int }_0^k\left(x^2\right)}dy\\ V=\pi {{\displaystyle {\int }_0^k\left(\sqrt{2y}\right)}}^{2}dy\\ V=\pi {\displaystyle {\int }_0^k\left(2y\right)}dy\\ V=2\pi {\left[\frac{y^2}{2}\right]}_0^k\\ V=\pi \left[k^2-0\right]\\ V=\pi k^2\end{array}$

ii.

To determine

To calculate: To solve a volume of the bowl generated by revolving the region.

Answer to Problem 22RE

  $V=\frac{1}{\pi }$ .

Explanation of Solution

Given information: The given information as follows:

a volume of the solid generated by revolving the region enclosed by the parabola $y=\frac{x^2}{2}$ and the from $y=0toy=2$ The bowl is filled at a rate of $2$ cubic units per second, when the depth $k$ is increasing when $k=1$ .

Calculation:

The volume of the solid generated by revolving the region bounded by the given parabola and the straight line can be calculated as:

Here, the depth $k$ increasing when $k=1$ .

Since, $V=\pi k^2$ .

  $\begin{array}{c}\frac{dV}{dt}=2\pi k\frac{dk}{dt}\\ \frac{dk}{dt}=\frac{1}{2\pi k}\frac{dV}{dt}\\ Ask=1,\frac{dV}{dt}=2\\ \frac{dk}{dt}=\frac{1}{2\pi \times 1}\times 2\\ \frac{dk}{dt}=\frac{1}{\pi }\end{array}$