Answer to Question #348182 in Calculus for Migz

Question

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

x=4 and the x-axis. What is the

the volume generated by revolving this

area about the y-axis?

Answer 1

$4^2=8y=>y=2$

$x^2=8y, x\ge0=>x=\sqrt{8y}$

$V=\pi \displaystyle\int_{0}^{2}(4^2-(\sqrt{8y})^2)dy$

$=\pi[16y-4y^2]\begin{matrix} 2\\ 0 \end{matrix}=\pi(32-16)$

$=16\pi ({units}^3)$