Solving Volume Problems

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Since the height of the prism is 5 units, there are 5 layers required to build the full prism, so the volume of the prism is 12 × 5 = 60 cubic units.\right|_0^h = \pi h\] So, we get the expected formula.Also, recall we are using \(r\) to represent the radius of the cylinder.What we need here is to get a formula for the cross-sectional area at any \(x\).In this case the cross-sectional area is constant and will be a disk of radius \(r\).You can see in the picture that the lateral faces are triangles, and that the edges of the lateral faces all meet at one point at the top, or vertex, of the pyramid.In this lesson, you solved problems involving the volume of rectangular pyramids and triangular pyramids.All of the examples in this section are going to be more general derivation of volume formulas for certain solids.As such we’ll be working with things like circles of radius \(r\) and we’ll not be giving a specific value of \(r\) and we’ll have heights of \(h\) instead of specific heights, .While \(r\) can clearly take different values it will never change once we start the problem.Cylinders do not change their radius in the middle of a problem and so as we move along the center of the cylinder ( the \(x\)-axis) \(r\) is a fixed number and won’t change.


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