How do you find the volume of the solid #y=4-x^2# revolved about the x-axis?

Question

Isaiah Allen · Accepted Answer

Use the disc method. The volume is #107.233#.

First, set #y# equal to #0# to find the bounds. #0=4-x^2# #x^2=4# #x in {-2,2}# So our bounds are #-2# and #2#. Next, use the disc method to find the volume (#y = r#). #int_-2^2piy^2 dx = piint_-2^2(4-x^2)^2dx# # = piint_-2^2(16-8x^2+x^4)dx# #= pi [16x-8/3x^3+x^5/5]_-2^2# #= pi (32-64/3+32/5)-pi(-32+64/3-32/5)# #=512/15pi# #=107.233# Final Answer

Emma Aldridge · Answer

undefined To find the volume of the solid generated by revolving the curve $y = 4 - x^2$ about the x-axis, you can use the method of cylindrical shells.

The volume $V$ is given by the integral:

$$ V = \int_{a}^{b} 2\pi x f(x) \, dx $$

where $ f(x) $ is the function defining the curve and $a$ and $b$ are the limits of integration.

For the given function $ y = 4 - x^2 $, we need to find the limits of integration, which are the x-values where the function intersects the x-axis. Since the function intersects the x-axis at $x = -2$ and $x = 2$, these will be our limits of integration.

So, the volume $V$ is:

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

You can then evaluate this integral to find the volume.