Answer to Question #227094 in Calculus for Simphiwe Dlamini

Question

The area bounded by the curves y=x²+2 and y=3x+2 from intersection point to intersection point is rotated around x=3. Determine the formula for the solid of revolution which is obtianed from this rotation. Explain your logic.

Answer 1

$$x^2+2=3x+2$$

$$x^2=3x$$

$$x_1=0, x_2=3$$

$x=0, y=0+2=2, Point(0,2)$

$x=3, y=3^2+2=11, Point(3,11)$

$$y=x^2+2, 0\leq x\leq 3=>x=\sqrt{y-2}, 2\leq y\leq 11$$

$$y=3x+2, 0\leq x\leq 3=>x=\dfrac{y-2}{3}, 2\leq y\leq 11$$

$$V=\pi\displaystyle\int_{2}^{11}((outer\ radius)^2-(inner\ radius)^2)dy$$

$$=\pi\displaystyle\int_{2}^{11}((3-\dfrac{y-2}{3})^2-(3-\sqrt{y-2})^2)dy$$

$$=\pi\displaystyle\int_{2}^{11}(\dfrac{121}{9}-\dfrac{22}{9}y+\dfrac{1}{9}y^2-9+6\sqrt{y-2}-y+2)dy$$

$$=\pi\displaystyle\int_{2}^{11}(\dfrac{1}{9}y^2-\dfrac{31}{9}y+\dfrac{58}{9}+6\sqrt{y-2})dy$$

$$=\pi\bigg[\dfrac{1}{27}y^3-\dfrac{31}{18}y^2+\dfrac{58}{9}y+4(y-2)^{3/2}\bigg]\begin{matrix} 11 \\ 2 \end{matrix}$$

$$=\dfrac{27\pi}{2}(units^3)$$