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# Find the volume of the described solid $S$.A right circular cone with height $h$ and base radius $r$.

## $V=\frac{1}{3} \pi r^{2} h$

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Applications of Integration

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suppose you want to find the volume of a right circular cone with height H. And base radius R. To do this we draw on the partition plane R. Right triangle with high H. And base radius R. Now looking at this, we see that the triangle is made up of the line. Y. Equal to H. And this line whose points are this would be Rh And this will be the origin 00. And so the lion formed by these two points would be why equal to that's a rice overrun of H. Over our. And then times X. Now if we rotate this right triangle about the why access can then form the right circular cone. And so using a strip that is parallel to the axis of rotation, the solid form by rotating this trip about the Y axis, it's a cylinder with radius equal to X. And height. That is the difference between the top function H. And the bottom function H. Over our X. And so by cylindrical shell method we have V equal to two pi times the integral from 0 to our. Since our area is found over the X interval zero to our and then you have radius which is X times height which is H minus H over our X. And then dx. And then simplifying we have two pi Integral from 0 to our of X. H minus H. Over our X squared dx. And so integrating with respect to X we have two pi times We have aged times x squared over two minus H. Over our times X cube over three. This evaluated from 0 to our So we have two pi times When exes are we have age times R squared over two minus H over our times are cube over three and when X zero everything becomes zero. So from here we have two pi times each over two R squared minus H over three R squared. And this is just two pi times each over six times R squared. Or this is just by over three H times R squared. Therefore the volume of the right circular cone is by over three R squared times each.

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Applications of Integration

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