mr. hanff needs to find the volume v of the solid obtained by rotating the region bounded by the curves…

mr. hanff needs to find the volume v of the solid obtained by rotating the region bounded by the curves about the specified dash line. you need to help him. set up the integral and find the volume by using the washer disk or washer method and by using the cylindrical shell method. you must set up and show all steps to reach the final answer.\n(a) set up the integral and find the volume using the disk or washer method.\n(b) set up the integral and find the volume using the cylindrical shell method.\nyou need to enter your setup and answer for both part (a) and (b). click insert > equation\ny = x²\nx = y²

mr. hanff needs to find the volume v of the solid obtained by rotating the region bounded by the curves about the specified dash line. you need to help him. set up the integral and find the volume by using the washer disk or washer method and by using the cylindrical shell method. you must set up and show all steps to reach the final answer.\n(a) set up the integral and find the volume using the disk or washer method.\n(b) set up the integral and find the volume using the cylindrical shell method.\nyou need to enter your setup and answer for both part (a) and (b). click insert > equation\ny = x²\nx = y²

Answer

Explanation:

Step1: Find intersection points

Set $x^{2}=\sqrt{x}$, then $x^{4}-x = 0$, $x(x^{3}-1)=0$. So $x = 0$ and $x = 1$ are the intersection - points.

Step2: Disk/Washer method (rotating about $y = 1$)

The outer - radius $R=1 - x^{2}$ and the inner - radius $r = 1-\sqrt{x}$. The volume formula for the washer method is $V=\pi\int_{a}^{b}(R^{2}-r^{2})dx$. Here, $a = 0$, $b = 1$. [ \begin{align*} V&=\pi\int_{0}^{1}[(1 - x^{2})^{2}-(1-\sqrt{x})^{2}]dx\ &=\pi\int_{0}^{1}(1 - 2x^{2}+x^{4}-(1 - 2\sqrt{x}+x))dx\ &=\pi\int_{0}^{1}(1 - 2x^{2}+x^{4}-1 + 2\sqrt{x}-x)dx\ &=\pi\int_{0}^{1}(x^{4}-2x^{2}-x + 2x^{\frac{1}{2}})dx \end{align*} ] [ \begin{align*} V&=\pi\left[\frac{x^{5}}{5}-\frac{2x^{3}}{3}-\frac{x^{2}}{2}+\frac{4x^{\frac{3}{2}}}{3}\right]_{0}^{1}\ &=\pi\left(\frac{1}{5}-\frac{2}{3}-\frac{1}{2}+\frac{4}{3}\right)\ &=\pi\left(\frac{6 - 20 - 15+40}{30}\right)\ &=\frac{11\pi}{30} \end{align*} ]

Step3: Cylindrical - shell method

We need to express the functions in terms of $y$. The curves are $x=\sqrt{y}$ and $x = y^{2}$. The volume formula for the cylindrical - shell method when rotating about $y = 1$ is $V = 2\pi\int_{c}^{d}(1 - y)(\sqrt{y}-y^{2})dy$, where $c = 0$, $d = 1$. [ \begin{align*} V&=2\pi\int_{0}^{1}(1 - y)(y^{\frac{1}{2}}-y^{2})dy\ &=2\pi\int_{0}^{1}(y^{\frac{1}{2}}-y^{2}-y^{\frac{3}{2}}+y^{3})dy\ &=2\pi\left[\frac{2y^{\frac{3}{2}}}{3}-\frac{y^{3}}{3}-\frac{2y^{\frac{5}{2}}}{5}+\frac{y^{4}}{4}\right]_{0}^{1}\ &=2\pi\left(\frac{2}{3}-\frac{1}{3}-\frac{2}{5}+\frac{1}{4}\right)\ &=2\pi\left(\frac{40 - 20 - 24 + 15}{60}\right)\ &=\frac{11\pi}{30} \end{align*} ]

Answer:

(a) The integral using the disk/washer method is $V=\pi\int_{0}^{1}(x^{4}-2x^{2}-x + 2x^{\frac{1}{2}})dx$ and the volume is $\frac{11\pi}{30}$. (b) The integral using the cylindrical - shell method is $V = 2\pi\int_{0}^{1}(y^{\frac{1}{2}}-y^{2}-y^{\frac{3}{2}}+y^{3})dy$ and the volume is $\frac{11\pi}{30}$.