shamieh
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Need someone to verify that my work is correct please. 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.
$$V = \int \pi (sin(x))^2 \, dx$$
$$ \pi \int^{\pi}_0 sin^2x \, dx$$
$$\pi \int^{\pi}_0 \frac{(1 - cos2x)}{2} \, dx$$
$$ \frac{1}{2} \pi \int^{\pi}_0 cos2x \, dx$$
$$\frac{1}{4} \pi \int^{\pi}_0 cos(u) \, dx$$
$$ u = 2x$$
$$\frac{du}{2} = dx$$
After updating the limits I get $\int^0_0$ thus: $0$
So I'm left with $$\frac{1}{4}\pi$$
a) Find the volume if the region is rotated about the x - axis.
$$V = \int \pi (sin(x))^2 \, dx$$
$$ \pi \int^{\pi}_0 sin^2x \, dx$$
$$\pi \int^{\pi}_0 \frac{(1 - cos2x)}{2} \, dx$$
$$ \frac{1}{2} \pi \int^{\pi}_0 cos2x \, dx$$
$$\frac{1}{4} \pi \int^{\pi}_0 cos(u) \, dx$$
$$ u = 2x$$
$$\frac{du}{2} = dx$$
After updating the limits I get $\int^0_0$ thus: $0$
So I'm left with $$\frac{1}{4}\pi$$