Question

Find the volume of the solid generated when the region bounded by the graph of \(y=\sin x\) and the \(x\) -axis on the interval \([0, \pi]\) is revolved about the \(x\) -axis.

Short answer

Question: Compute the volume of the solid generated when the region bounded by the graph of \(y=\sin x\), the \(x\)-axis, and the interval \([0, \pi]\) is revolved about the \(x\)-axis. Answer: The volume of the solid is \(\frac{\pi^2}{2}\).

Step-by-step solution

Set up the volume integral

Write the formula for the volume of the solid when the region is revolved about the \(x\)-axis. For this problem, we have \(f(x) = \sin x\), \(a=0\), and \(b=\pi\). Plugging these values into the formula, we get: $$V = \int_0^{\pi} \pi (\sin x)^2 dx$$

Integrate the function

We need to integrate \(\pi (\sin x)^2\) with respect to \(x\) over the interval \([0, \pi]\). To do this, we can use the double-angle formula: $$2 \sin^2 x = 1-\cos 2x$$ $$\sin^2 x = \frac{1-\cos 2x}{2}$$ Now, replace \((\sin x)^2\) with this expression in the integral: $$V = \int_0^{\pi} \pi \left(\frac{1-\cos 2x}{2}\right) dx$$ Next, distribute \(\pi\), and break the integral into two separate integrals: $$V = \int_0^{\pi} \frac{\pi}{2} dx - \int_0^{\pi} \frac{\pi \cos 2x}{2} dx$$ Now integrate each part separately: $$V = \frac{\pi}{2} \int_0^{\pi} dx - \frac{\pi}{2} \int_0^{\pi} \cos 2x dx$$ $$V = \frac{\pi}{2} (x \Bigr |_{0}^{\pi}) - \frac{\pi}{2} \left(\frac{\sin 2x}{2} \Bigr |_{0}^{\pi}\right)$$

Evaluate the integrals

Evaluate the two integrals at the bounds, and then simplify the expression: $$V = \frac{\pi}{2} (\pi - 0) - \frac{\pi}{2} \left(\frac{\sin 2\pi}{2} - \frac{\sin 0}{2}\right)$$ Since \(\sin 2\pi = 0\) and \(\sin 0 = 0\), the second term becomes zero: $$V = \frac{\pi}{2} (\pi - 0) - \frac{\pi}{2}(0)$$ Therefore, the volume of the solid is: $$V = \frac{\pi^2}{2}$$