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# The base of $S$ is a circular disk with radius $r$. Parallel cross-sections perpendicular to the base are isosceles triangles with height $h$ and unequal side in the base. (a) Set up an integral for the volume of $S$. (b) By interpreting the integral as an area, find the volume of $S$.

## a) $V=2 h \int_{0}^{r} \sqrt{r^{2}-x^{2}} d x$b) $V=\frac{1}{2} \pi r^{2} h$

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Applications of Integration

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So for this problem, we have a circle with radius R, and it's centered at the origin. So we have is X squared, plus y squared equals r squared. Um, and we solve for y and we see that y equals plus or minus the square root of R squared minus X squared. So we know that the top half will be the positive square root. Um, and the bottom half will be the negative square root. So the base that we have is two times r squared, minus x squared, um, and that will give us the whole base length of RSO sleeves triangle, Then the height is the same for every triangle. So the area of one cross section is going to be FX equal to one half our base, which is to route R squared minus X squared. Times are height, which is going to equal our height, times the square root. Hi, because the tubes will cancel. So just high times square r squared my sex work. Then we know the circle intersects at the X. The X axis at X equals negative. R and X equals positive are so that's how we're going to integrate we can pull out the H, and we also know that it's symmetric. So rather than going from negative, are are we can just go from zero to our and multiply it by two. So our volume is going to equal two times h times the interval from zero to our of the R squared minus X squared G x then for part B. Um, what we find is using this. We can take the we can square both sides. We get our equation of a circle again. Um so since Y equals the square root of X r squared minus X squared, that's our upper half of the circle of radius R centered at the origin. So the way that we can write this now is going to be the volume once again, up to age, we have that same thing. But that's actually gonna end up equaling two h times. One fourth of pi r squared. So then the final answer will be one half pi r squared H and we were able to do this by replacing the integral with the area of a circle which we know to be pi r squared. And then we multiply that by 1/4. Since we have, um, half of the half circle

California Baptist University

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Applications of Integration

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