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Consider the region bounded by $y = sin(x)$ and the x - axis from $ x = 0$ to $x = \pi$

a) Find the volume if the region is rotated about the x - axis.

\(\displaystyle V = \int \pi (sin(x))^2 \, dx\)

\(\displaystyle \pi \int^{\pi}_0 sin^2x \, dx\)

\(\displaystyle \pi \int^{\pi}_0 \frac{(1 - cos2x)}{2} \, dx\)

\(\displaystyle \frac{1}{2} \pi \int^{\pi}_0 cos2x \, dx\)

\(\displaystyle \frac{1}{4} \pi \int^{\pi}_0 cos(u) \, dx\)

\(\displaystyle u = 2x\)

\(\displaystyle \frac{du}{2} = dx\)

After updating the limits I get $\int^0_0$ thus: $0$

So i'm left with \(\displaystyle \frac{1}{4}\pi\)