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Calculus/Pool Volume

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Expert: - 2/27/2010


Question
Consider the region bounded by f(x)=sin(pi x) and g(x)=x^3-4x.
The given region is the surface of a pool with depth given by
h(x)=sin((pi/2)x). Find the volume of the pool.

Answer
The curves f(x) and g(x) intersect at -2, 0, and 2.
Since we are talking about x as a measurement, I'll ignore the -2 since distance is non-zero.
Thus, we have an integral from 0 to 2 of f(x) - g(x), but it is also times the depth.

The full equation would be ∫ (sin(πx) - x³ + 4x)sin(πx/2) dx for x = 2 down to 10.

This can be multiplied out and broken into ∫ sin(πx)sin(πx/2) dx - ∫x³sin(πx/2) dx + ∫ 4x*sin(πx/2) dx.

Note that sin(πx) = 2sin(πx/2)cos(πx/2).
This makes the 1st integral a u substitution with u = sin(πx/2).

For the 2nd integral, use a u-dv substitution 3 times, remembering there is a minus sign out front.
In the 1st u-v integral, u = x³ and dv = sin(πx/2)dx.
In the 2nd u-v integeral, u = x² and dv = sin(πx/2)dx.

At this point, combine the integral with the 3rd one given up above,
since they both have k*x*sin(πx) for some constant k.
For this combined integral, put the constant out front and let u = x  and dv = sin(πx)dx.
This should put it in a from that can be easily done with some consants out front.

Remember to evaluate the last integral and all the constants from 0 to 2.

That should be enough to get the problem done, but I will provide more help if requested.

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