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Find the volume, on the interval 0\leq x\leq 3, of a 3-D object whose cross-section at any given point is a square with side length x^{2}-9x.

First, think about finding the area under a curve. We integrate the function to do this but essentially what we are doing is adding up an infinite amount of infinitely skinny lines and these lines (1-D objects) add up to give us the area. The function we integrate gives us the length of the line. It is a generalization of the formula A=l\times h for a rectangle. We get the infinitely many heights and we add them up over the length, the difference being that they vary. This problem is very similar though it might seem very different. Here, we are adding up an infinite amount of infinitely thin squares (2-D objects) and we will end up with the volume of the object. The function we integrate will give us the area, which we then add up to get a volume. Thus, the formula for finding a volume by cross-sectional areas is

V=\int _{{a}}^{{b}}A(x)\,dx\,

where we are finding the volume on the interval [a,b] and where A(x) gives a formula for the area of a general cross-section. Here, we have

V=\int _{{0}}^{{3}}(x^{2}-9x)^{2}\,dx\,

because the area of a square is equal to the square of its side length.

V=\int _{{0}}^{{3}}(x^{2}-9x)^{2}\,dx=\int _{{0}}^{{3}}(x^{4}-18x^{3}+81x^{2})\,dx=\left[{\frac  {1}{5}}x^{5}-{\frac  {9}{2}}x^{4}+27x^{3}\right]_{{0}}^{{3}}={\frac  {4131}{10}}=413.1\,

So, the volume of this shape, whose cross-section is always a square with varying side length, is 413.1 cubic units.

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