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

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 disk - washer method (in terms of y)

The outer radius $R(y)=(-7)-(-6 + y^{2})=1 - y^{2}$, and the inner radius $r(y)=(-7)-(-y)=y - 7$. The volume formula is $V=\pi\int_{a}^{b}(R^{2}(y)-r^{2}(y))dy$, where $a=-3$ and $b = 2$. $V=\pi\int_{-3}^{2}((1 - y^{2})^{2}-(y - 7)^{2})dy$.

Step3: Expand the integrands

$(1 - y^{2})^{2}=1-2y^{2}+y^{4}$ and $(y - 7)^{2}=y^{2}-14y + 49$. So $(1 - y^{2})^{2}-(y - 7)^{2}=1-2y^{2}+y^{4}-(y^{2}-14y + 49)=y^{4}-3y^{2}+14y - 48$.

Step4: Integrate term - by - term

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

Step5: Evaluate the definite integral

$V=\pi\left[\frac{y^{5}}{5}-y^{3}+7y^{2}-48y\right]_{-3}^{2}$. $V=\pi\left[\left(\frac{2^{5}}{5}-2^{3}+7\times2^{2}-48\times2\right)-\left(\frac{(-3)^{5}}{5}-(-3)^{3}+7\times(-3)^{2}-48\times(-3)\right)\right]$. $V=\pi\left[\left(\frac{32}{5}-8 + 28-96\right)-\left(-\frac{243}{5}+27 + 63 + 144\right)\right]$. $V=\pi\left[\left(\frac{32}{5}-76\right)-\left(-\frac{243}{5}+234\right)\right]$. $V=\pi\left[\frac{32}{5}-76+\frac{243}{5}-234\right]$. $V=\pi\left[\frac{32 + 243}{5}-310\right]$. $V=\pi\left[\frac{275}{5}-310\right]$. $V=\pi(55 - 310)$. $V = 255\pi\approx799.529$.

Answer:

$799.529$