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

Explanation:

Step1: Find intersection points

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

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

The radius of a shell is $r=(x+7)$ and the height of the shell is $h=(-y)-(-6 + y^{2})=-y^{2}-y + 6$. The volume formula for rotating about a vertical line $x = a$ using cylindrical - shells is $V = 2\pi\int_{c}^{d}r\cdot hdy$. Here, $c=-3$, $d = 2$, and $r=(y + 7)$. So $V=2\pi\int_{-3}^{2}(y + 7)(-y^{2}-y + 6)dy$.

Step3: Expand the integrand

$(y + 7)(-y^{2}-y + 6)=-y^{3}-y^{2}+6y-7y^{2}-7y + 42=-y^{3}-8y^{2}-y + 42$.

Step4: Integrate term - by - term

$\int(-y^{3}-8y^{2}-y + 42)dy=-\frac{1}{4}y^{4}-\frac{8}{3}y^{3}-\frac{1}{2}y^{2}+42y+C$.

Step5: Evaluate the definite integral

$V = 2\pi\left[-\frac{1}{4}y^{4}-\frac{8}{3}y^{3}-\frac{1}{2}y^{2}+42y\right]_{-3}^{2}$. First, substitute $y = 2$: $-\frac{1}{4}(2)^{4}-\frac{8}{3}(2)^{3}-\frac{1}{2}(2)^{2}+42(2)=-4-\frac{64}{3}-2 + 84=\frac{-12-64 - 6+252}{3}=\frac{170}{3}$. Then substitute $y=-3$: $-\frac{1}{4}(-3)^{4}-\frac{8}{3}(-3)^{3}-\frac{1}{2}(-3)^{2}+42(-3)=-\frac{81}{4}+72-\frac{9}{2}-126=\frac{-81 + 288-18 - 504}{4}=\frac{-315}{4}$. $V = 2\pi\left(\frac{170}{3}-\left(-\frac{315}{4}\right)\right)=2\pi\left(\frac{680 + 945}{12}\right)=2\pi\times\frac{1625}{12}=\frac{1625\pi}{6}\approx850.805$.

Answer:

$850.805$