Topic: solids of revolution problem (Read 388 times)
I think I do. Thank for your response!
Because the integral of a curve is equal to the area under it, which has obvious implications when it comes to geometry. The integral ∫ 2f(x) dx is the area of the solid's cross section. The volume of the solid can be calculated by π ∫ f(x)2 dx.
Consider the simplest case, a cylinder (f(x)=r). The area of the cross section (a rectangle) (∫ 2r dx) is 2r times the length, and the volume (π ∫ r2 dx) is πr2 times the length.
Now consider a cone whose base radius is equal to its length (f(x)=x). The area of the cross section (a triangle) (∫ 2x dx) is the square of the length, and the volume (π ∫ x2 dx) is naturally πx3/3.
Does that answer your question?
Consider the simplest case, a cylinder (f(x)=r). The area of the cross section (a rectangle) (∫ 2r dx) is 2r times the length, and the volume (π ∫ r2 dx) is πr2 times the length.
Now consider a cone whose base radius is equal to its length (f(x)=x). The area of the cross section (a triangle) (∫ 2x dx) is the square of the length, and the volume (π ∫ x2 dx) is naturally πx3/3.
Does that answer your question?
So I can't seem to understand why the width of anew infinitesimal cylinder is dx. Where does it come from? What does in have to do with the integral?
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