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

Explanation:

Step1: Find intersection points

Set $-6 + y^{2}=y$, then $y^{2}-y - 6=0$. Factoring gives $(y - 3)(y+2)=0$. So $y=-2$ and $y = 3$.

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

The radius of a shell is $r=(x + 9)$ and the height of the shell is $h=(-6 + y^{2}-y)$. The volume formula for the method of cylindrical - shells about a vertical line $x = a$ is $V=2\pi\int_{c}^{d}r\cdot hdy$. Here, $c=-2$, $d = 3$, $r=(y + 9)$ (since $x=y$ is the right - hand curve and $x=-6 + y^{2}$ is the left - hand curve), and $h=(-6 + y^{2}-y)$. So $V=2\pi\int_{-2}^{3}(y + 9)(-6 + y^{2}-y)dy$. Expand the integrand: $(y + 9)(-6 + y^{2}-y)=y(-6 + y^{2}-y)+9(-6 + y^{2}-y)=-6y+y^{3}-y^{2}-54 + 9y^{2}-9y=y^{3}+8y^{2}-15y - 54$.

Step3: Integrate

$\int(y^{3}+8y^{2}-15y - 54)dy=\frac{y^{4}}{4}+\frac{8y^{3}}{3}-\frac{15y^{2}}{2}-54y+C$.

Step4: Evaluate the definite integral

$V = 2\pi\left[\frac{y^{4}}{4}+\frac{8y^{3}}{3}-\frac{15y^{2}}{2}-54y\right]_{-2}^{3}$ $=2\pi\left[\left(\frac{3^{4}}{4}+\frac{8\times3^{3}}{3}-\frac{15\times3^{2}}{2}-54\times3\right)-\left(\frac{(-2)^{4}}{4}+\frac{8\times(-2)^{3}}{3}-\frac{15\times(-2)^{2}}{2}-54\times(-2)\right)\right]$ $=2\pi\left[\left(\frac{81}{4}+72-\frac{135}{2}-162\right)-\left(4-\frac{64}{3}-30 + 108\right)\right]$ $=2\pi\left[\left(\frac{81}{4}+72-\frac{270}{4}-162\right)-\left(4-\frac{64}{3}+78\right)\right]$ $=2\pi\left[\left(\frac{81 - 270}{4}+72-162\right)-\left(82-\frac{64}{3}\right)\right]$ $=2\pi\left[\left(-\frac{189}{4}-90\right)-\left(\frac{246 - 64}{3}\right)\right]$ $=2\pi\left[-\frac{189}{4}-90-\frac{182}{3}\right]$ $=2\pi\left[-\frac{189\times3+90\times12 + 182\times4}{12}\right]$ $=2\pi\left[-\frac{567+1080+728}{12}\right]$ $=2\pi\left[-\frac{2375}{12}\right]$ $=\frac{2\pi\times2375}{12}=\frac{2375\pi}{6}\approx1243.471$.

Answer:

$1243.471$