Answer to Question #221524 in Calculus for cayyy

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 the solid of revolution which is obtained from this rotation.

Answer 1

The volume of the solid bounded by $y = x^2 + 2 \space and \space y = 3x + 2$ about x-axis

$x^2 + 2 = 3x + 2 \\ x^2-3x=0\\ x=0,3\\ Volume = \int_0^3 \pi [(3x+2)^2-(x^2+2)^2]dx\\$

The volume of the solid bounded by $y = x^2 + 2 \space and \space y = 3x + 2$ about x=3

$x^2 + 2 = 3x + 2 \\ x^2-3x=0\\ x=0,3\\ x= 0\implies y= 2\\ x= 3 \implies y= 11\\ Volume = \pi \int_2^{11} \pi [(R(x))^2-(r(x))^2]dx\\ y= 3x+2 \implies x= \frac{y-2}{3}\\ y=x^2+2 \implies x= \sqrt{y-2}\\ R(x) = 3- (\frac{y-2}{3})\\ r(x)= 3-\sqrt{y-2}\\ Volume = \pi \int_2^{11} \pi [(3- (\frac{y-2}{3}))^2-(3-\sqrt{y-2})^2]dx\\$