Find the volume of the solid obtained when the region under the curve $ y = \arcsin x $, $ x \ge 0 $, is rotated about the y-axis.
Okay, This question Once a stir of all of the area underneath Weikel sine inverse X, where X is greater than zero around the y axis. So, with all volume of revolution questions, let's draw to see what we're dealing with. So Arc Sine has a domain from negative one No one. So if we're considering the positive side, we end at one here. And then if we graph park sign of X, it looks something like this, but we're evolving this area around the Y axis. So what we can do is convert this to a function of X and consider this area right here. And remember this Y value is pirate, too, because arc sine of one is pirates. So to convert to a function of acts, we could just see that sign of why equals X. So that becomes are integral. So are equals. Sign Why? So that means air volume is pi times the integral from zero to pi over too of sign of why squared d y and no, that's an integral that we can look up. Or at this point, you may have it memorized. So the integral of sine squared is 1/2 why minus 1/4 sign of two. Why? From 0 to 2. Pi Sorry, zero pi over too. And this becomes pi Times pi over two divided by two minus Well signed pie is zero minus 1/2 time. Zero minus 1/4 time. Zero. So this is just equal to pi times pi over two divided by two, which is equal to high times over for or pi squared over for so that's our final answer.