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Numerade Educator



Problem 60 Hard Difficulty

Find the volume of the described solid $ S $.
The base of $ S $ is the region enclosed by $ y = 2 - x^2 $ and the x-axis. Cross-sections perpendicular to the y-axis are quarter-circles.


$V=2 \pi$


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Video Transcript

were given a solid S. And we're asked to find the volume of this solid Dude, we're told the base of S. Is the region enclosed by the curves? Y equals two minus X squared. And the X axis on the tip of the chain wallet chain. And we're told that the cross sections perpendicular to the y axis are quarter circles. Listen, so first you want to find the volume. We need to find the area of each of these quarter circles. To find the area of the quarter circles need to find the radius of them to find the radius. We need to find the distance between the left and right sides of our region. Well, solving for X, we get that X is equal to plus or minus the square root of two minus Y. And so therefore we have the radius go of each of our quarter circles. I'm sorry, is actually going to be using symmetry two times the square root of two minus Y. Didn't have to say. And it's adam do you said it? Mhm. And therefore we have that the area of the quarter circle. Well this is 1/4 of pi r squared, Which is 1 4th times pi times two times route to minus Y squared. This is the same as two -Y. Ooh some times pi mm somebody goes now, since this is the area of the quarter circle as a function of why we have their volume. This is the integral over all these quarter circles. So we see why ranges from zero up to the vertex of the parabola too. And this is integral from 0 to 2 of a of Y DY, Which is pi times the integral from 0 to 2 of two -Y. Do you why evaluating this integral? We get two pi. This is our answer.