find the volume of the solid obtained by rotating the region bounded by x = 6 - y² and x = -y about the line…

Explanation:

Step1: Find intersection points

Set $6 - y^{2}=-y$. Rearrange to $y^{2}-y - 6=0$. Factor: $(y - 3)(y+2)=0$. So $y=-2$ and $y = 3$.

Step2: Use the method of cylindrical - shells (in terms of y)

The radius of a shell is $r=7 - x$. For $x = 6 - y^{2}$, $r_1=7-(6 - y^{2})=1 + y^{2}$; for $x=-y$, $r_2=7 + y$. The height of the shell $h=(6 - y^{2})-(-y)=6 - y^{2}+y$. The volume formula using the method of cylindrical - shells about the line $x = 7$ (in terms of y) is $V=\pi\int_{a}^{b}[(r_1^{2}-r_2^{2})]dy$. Here, $V=\pi\int_{-2}^{3}[(1 + y^{2})^{2}-(7 + y)^{2}]dy$.

Step3: Expand the integrand

Expand $(1 + y^{2})^{2}=1 + 2y^{2}+y^{4}$ and $(7 + y)^{2}=49+14y + y^{2}$. Then $(1 + 2y^{2}+y^{4})-(49+14y + y^{2})=y^{4}+y^{2}-14y - 48$.

Step4: Integrate

$\int(y^{4}+y^{2}-14y - 48)dy=\frac{y^{5}}{5}+\frac{y^{3}}{3}-7y^{2}-48y+C$.

Step5: Evaluate the definite integral

$V=\pi\left[\frac{y^{5}}{5}+\frac{y^{3}}{3}-7y^{2}-48y\right]_{-2}^{3}$. $V=\pi\left[\left(\frac{3^{5}}{5}+\frac{3^{3}}{3}-7\times3^{2}-48\times3\right)-\left(\frac{(-2)^{5}}{5}+\frac{(-2)^{3}}{3}-7\times(-2)^{2}-48\times(-2)\right)\right]$. $V=\pi\left[\left(\frac{243}{5}+9 - 63-144\right)-\left(-\frac{32}{5}-\frac{8}{3}-28 + 96\right)\right]$. $V=\pi\left[\left(\frac{243}{5}-198\right)-\left(-\frac{32}{5}+68-\frac{8}{3}\right)\right]$. $V=\pi\left[\frac{243}{5}-198+\frac{32}{5}-68+\frac{8}{3}\right]$. $V=\pi\left[\frac{243 + 32}{5}-(198 + 68)+\frac{8}{3}\right]$. $V=\pi\left[\frac{275}{5}-266+\frac{8}{3}\right]$. $V=\pi\left[55-266+\frac{8}{3}\right]$. $V=\pi\left[-211+\frac{8}{3}\right]$. $V=\pi\left[\frac{-633 + 8}{3}\right]=\frac{\pi(-625)}{3}\approx - 654.498$. Since volume is non - negative, $V\approx654.498$.

Answer:

$654.498$