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# Find the volume of the described solid $S$.The base of $S$ is the region enclosed by the parabola $y = 1 - x^2$ and the x-axis. Cross-sections perpendicular to the y-axis are squares.

## 2 units $^{3}$

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Applications of Integration

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were given a solid S and rest to find the volume of this solid, we're told the base of S. Is the region enclosed by the parabola. Y equals one minus X squared and the X. Axis and the cross section is perpendicular to the y axis are squares. That's something good. What does he said have a good woman that I've never seen? Well to find the volume will find the area of each of these squares. To find the area of the square. As you want to find the length of one of these sides of the squares, which is the distance from one side of the parabola to the other. To do this, we want to solve for X as a function of Y. So we have rearranging X equals mm plus or minus the square root of one minus Y. And therefore it follows their side length. S is going to be two times the square root of 1-. Why. Yeah. And therefore our area of the square as a function of why? Well this is S squared Which is four times 1 -Y. Really. And now using the volume interpretation uh Benadryl, the volume is the integral from why ranges. Yeah. More of the purest I think. Yeah, I've never seen it sort of the from zero the x axis up to the Vertex of the Parabola one of the area A. Of Y. Dy. Would you think? Mhm Towards you harry If you evaluate this integral, this is the integral from 0 to 1 of four times one minus Y. Dy. Which is equal to two observe this. Let's hear a brand new tech and

Ohio State University

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Applications of Integration

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