Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.
$ y^2 = x $ , $ x = 2y $ ; about the y-axis
Applications of Integration
we're finding the volume of the solid obtained by rotating the region. Bounded by Why score equals X X equals two. Why rotating about the y axis? So let's first start off by station in the region. So this is the X and Y axes. We can first start off by doing that. Why squared equals X. So the wind was X squared Line typically looks like this. Yeah, but when we take the inverse of that, we basically you can't look like this because no matter what the value of why is X is always positive. We're basically rotating by ninety degrees clockwise, and then we can do the what X equals to wildlife. So normally X equals Y looks like this, but we're stretching it out a little bit more, you know that. So now when you find the two points of intersection So the very clear one is your zero and the other one we can solve by recording these two equations together. So we get why squared is equal to two. Why? And the very obvious answer here is that y equals two. Because if why times wise equal to two times alive. And why has to be two. And if we plugged that in and we get X equals four. So this point right here where the two lines intersect is for two. We're voting this all about the y axis for here. Oh, this is the region that we care about. So now when you draw out, the typical Washington is created, and so it looks like a washer. A Samantha circles attracted out from the marker circle. So what do you find? The cross sectional area that we find in terms of why this is Why changes the area changes. The bigger circle is caused by this line. Right here is that school is always the exam is always greater, creating the larger circle. So that's X equals to why we plug in, um, the axe value because the X value creates the is the radius of the circle. So the big R is just too wise, and so therefore the smaller has to be just y squared. And if we finally simplify all of this, you get that's equal to pi comes for squared. Mine's wants before and so that is our cross sectional area function. We can now go ahead and proceed with our interval. So as I mentioned before, for two is the point of intersection along with origin. So we're going from Wankel zero to y equals two. When we put in the cross sectional area function that we found before running merely plowed the part and then was left is for my square minus y to the fourth d Y, and we can find this too anti differentiation. We said that this is pies the anti derivative of four. Why square is just four times why three to over three minus y five over five. That's the anti driven apply to the forth that is a valid way to add zero and two. We can plug this end. We get it. Pi times for two to the three is eight three minus two to the fifth is thirty two or sixteen? No, actually, it's Terry to five. We don't care about the case where X equals zero or Y equals zero, because it just gives us hero. So this is our final answer. We multiply all this out when you finally get the answer. That's this Entire Integral is equal to sixty four pi over fifteen