Find the volume of the solid obtained by rotating the region bounded by the curves.

Question

Find the volume of the solid obtained by rotating the region bounded by the curves y=tan^2(x), x=pi/4, x=0, and y=0 about the x-axis. PLEASE HELP ASAP.

Answer 1

You would get a series of circles or "disks", each with an area of πr² so the volume of each would be πr²•dx with the radius being the "y" value. So, since y = tan²(x), the total volume is:

₀∫^π/4 π(tan²(x))² dx = π ₀∫^π/4 tan⁴(x) dx

To calculate the antiderivative, see Doug C's response in the comment to your question. Otherwise, just use your calculator to get the answer.

Answer 2

I am assuming the region is between the tan^2(x) and the x-axis (y = 0). The x = 0 doesn't help bound anything.

Draw the region with a graphing calculator desmos.com

Slice the volume of revolution perpendicular to the x axis, into thin disks of thickness dx, radius y(x) = tan^2(x) and integrate for x = 0 to pi/4:

Volume = Integral of pi * y^2(x) * dx for x = 0 to pi/4.