Answer to Question #348183 in Calculus for Migz

Question

Given the area in the first quadrant bounded by x² = = 8y, the line

y-2 and the y-axis. What is the volume generated when this area is

revolved about the line y - 2 = 0?

Answer 1

$$y=2=>x^2=8(2)=>x=\pm4$$

$$V=\pi \displaystyle\int_{-4}^{4}(2-\dfrac{x^2}{8})^2dx$$

$$V=\pi \displaystyle\int_{-4}^{4}(4-\dfrac{x^2}{2}+\dfrac{x^4}{64}))dx$$

$$=\pi[4x-\dfrac{x^3}{6}+\dfrac{x^5}{320}]\begin{matrix} 4\\ -4 \end{matrix}$$

$$=\pi(16-\dfrac{64}{6}+\dfrac{1024}{320}-(-16+\dfrac{64}{6}-\dfrac{1024}{320}))$$

$$=\dfrac{256\pi}{15}({units}^3)$$