Question

Use the shell method to compute the volume of the solids obtained by rotating the region enclosed by the graphs of the functions y=x2 and y=8−x2 for x≥0.5 about the y-axis.

Answer 1

Answer:

44.18 cubic unit

Step-by-step explanation:

The intersection of the curve $y = x^2$ and $y = 8-x^2$ is

$x^2 = 8 - x^2$

$2x^2 = 8$

$x^2 = 4$

$x = \pm 2$

Since we are rotating the region where $x \geq 0.5$ we will use the intersection x = 2 and y = 4

Using the shell method with x ranges from 0.5 to 2. The volume of the shell would be

$V = \int\limits^2_{0.5} {2\pi x h} \, dx$

where h is the difference between the y-coordinates of the curves, in other words

$h = 8 - x^2 - x^2 = 8 - 2x^2$

Plug h into the V integral and we have

$V = \int\limits^2_{0.5} {2\pi x (8 - 2x^2)} \, dx\\V = 4\pi\int\limits^2_{0.5} {(4x - x^3)} \, dx\\V = 4\pi[2x^2 - x^4/4]^2_{0.5}\\V = 4\pi(2*2^2 - 2^4/4 - 2*0.5^2 + 0.5^4/4)\\V = 4\pi(8 - 4 - 0.5 + 0.015625) \approx 44.18$