The region under the curve $ y = \sin^2 x $ from 0 to $ \pi $ is rotated about the x-axis. Find the volume of the resulting solid.
Okay, This question wants us to find the volume of the solid formed by revolving the area under Weikel sine X squared from zero to pie around the X axis. So to do this, let's draw a picture and we're going to get something like this. So this is our area in question and zero is that pie. So we want to revolve this around the x axis so we can just go ahead and use the disc method with our radius equal to why which is equal to sine squared X. So now we can set up our integral for the volume. So we get pie times the integral from zero to pie of sine squared X squared the ex and that turns into pi times The answer girl from zero pie of Signed to the fourth ex d X. So now we have to use a table to find the integral of sine to the fourth. So we get pi times are inside Derivative which using a table we confined is three acts over eight plus 1/8 sign ex co sign cubed acts minus 1/8 Signed cubed co sign minus 1/2 sine x coastline X and we're evaluating this from zero to pie. And if we look here Sign of zero and sign of pie are both zero. So everything with a sign goes away. So that just means our answer is pi Times three pi over a minus zero or three. Pi squared over eight and that's our final answer for the volume.