Answer to Question #348510 in Calculus for Migz

Question

Given the area in the first quadrant bounded by y^2 = x, the line x =

4 and the x-axis. What is the volume generated when this area is

revolved about the y-axis?

Answer 1

The volume of a solid generated when the area bounded by the curve x = f(y) and the y-axis between

y = c and y = d is revolved about the y-axis:

$V=\pi \int_c^d [f(y)]^2 dy.$

So we have:

$V=\pi \int_0^2 [4]^2dy - \pi \int_0^2[y^2]^2dy=$

$=\pi \int_0^2 16dy - \pi \int_0^2y^4dy=$

$=\pi(16y-\frac{y^5}{5})|_0^2=\pi(16\cdot 2-\frac{2^5}{5})=\pi(32-\frac{32}{5})=$

$=\pi(32-6.4)=25.6\pi\approx 80.4$ cubic units.

Answer: $25.6\pi\approx 80.4$ cubic units.