Question

The region bounded by the given curve is rotated about the specified axis. Find the volume V of the resulting solid by any method. x2 + (y − 4)2 = 16;    about the y-axis

Answer 1

`\bf x^2+(y-4)^2=16\implies x^2=16-(y-4)^2 \\\\\\ x=√(16-(y-4)^2)\implies x=√(16-(y^2-8y+16)) \\\\\\ x=√(8y-y^2)\\\\ -------------------------------\\\\ 0=√(8y-y^2)\implies 0=8y-y^2\implies 0=y(8-y) \implies y= \begin{cases} 0\\ 8 \end{cases}\\\\ -------------------------------\\\\ \displaystyle \stackrel{\textit{disk method}}{\int\limits_(0)^8~\pi (√(8y-y^2))^2\cdot dy}\implies \pi \int\limits_(0)^8~ 8y-y^2\cdot dy`


`\bf \displaystyle \pi \int\limits_(0)^8~8y\cdot dy - \pi \int\limits_(0)^8~y^2\cdot dy\implies \pi \left[ 4y^2~-~\cfrac{y^3}{3} \right]_(0)^8 \\\\\\ \pi \left[ \left[ 256-\cfrac{51}{3} \right]~-~[0] \right]\implies \pi \cdot \cfrac{256}{3}\implies \cfrac{256\pi }{3}\implies 85(\pi )/(3)`