Find the volume of the described solid $ S $.

A frustum of a right circular cone with height $ h $, lower base radius $ R $, and top radius $ r $.

$\frac{\pi h}{3}\left(R^{2}+R r+r^{2}\right)$

Applications of Integration

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in order to solve this problem. The first step we can do is we can draw a graphical representation of it. And again, you can do this using does mouse. Okay, Now that we know this, we know the raid. I, in other words, multiple radius is, can be seen by looking at the formula. Why, minus y one as equipment, um, times x minus X one. This is also known this point So formula in algebra writing this in terms of capital and lower case R, which are the variables we're using rather than X are. Then why, in this context and x one, we have this. We cannot transition over to volume. We know that D backs is equivalent to H over our minus there capital First, because we know that you is capital R minus blockades are over h times. Experts are obviously we know that we just end up with the constants which give us a derivative of zero. So we have. This is our value for D X, which means that we can now write our final volume as high H over three times. Couple or Meisler case are we already drive this part and then in simplest form. Hi. H over three. Recognize that stuff can be crossed off quick. And remember this this answer It all came back from getting the point slope form and from getting our M.