finding the volume of a solid using shell method (calculus II)
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Imagine you have a circle. To calculate the area of the circle, you multiply the radius R by pi and 2. The radius R is the distance from the center of the circle to the edge. Simple.
Let’s give this circle some height/depth, so it’s a regular cylinder with height H. To find the volume, you multiply the surface area by the height, giving you R•H•2•pi. Still pretty simple.
Now let’s assume the height H isn’t constant, but changes as it goes outward from the center, like the tip of a bullet or a cone. We can represent the height H as a function of the distance from the center: H = f(x), where x goes from 0 (the center) to R (the outer edge). If we integrate H with respect to x from 0 to R, we get the area under the curve, which is int-H(x)•(R-0). This approximates the same overall value as the H•R term in the simple cylinder equation, which is the area of a slice of the cylinder. So continuing the approach from the cylinder, we multiply by 2pi to turn that single slice into a whole circular shape.
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