definite integrals

Apr 2005
\(\displaystyle \int_0^6 y(6y- y^2)^{1/2} dy\)

The first thing I would do is "complete the square" inside the square root:
\(\displaystyle \int_0^6 y(9- (y- 3)^2)^{1/2}dy\)

Let u= y- 3. Of course, y= u+ 3 and when y= 0, u=-3, when y= 6, y= 3. This is now \(\displaystyle \int_{-3}^3 (u+3)(9- u^2)^{1/2}du= u\int_{-3}^3 u(9- u^2)^{1/2}du+ 3\int_{-3}^3 (9- u^2)^{1/2}du\)

To integrate \(\displaystyle \int_{-3}^3 u(9- u^2)^{1/2} du\) let \(\displaystyle v= 9- u^2\).

To integrate \(\displaystyle 3\int_{-3}^3 (9- u^2)^{1/2}du\) let \(\displaystyle u= 3 sin(\theta)\).
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