Question

Find the volume of the solid generated when the region bounded by \(y=\cos x\) and the \(x\) -axis on the interval \([0, \pi / 2]\) is revolved about the \(y\) -axis.

Short answer

Answer: The volume of the solid is \(\frac{\pi^2}{2}\).

Step-by-step solution

Set up the integral

To set up the integral, we first need to determine the inequalities that bound the region. The region is bounded by the curve \(y=\cos x\), the x-axis, and the range \([0, \pi / 2]\). As the solid is revolved about the \(y\)-axis, the radius of each washer will be a function of \(y\). We notice that the distance from the \(y\)-axis to the curve \(y=\cos x\) is just \(x\), which on our region goes from \(0\) to \(\pi / 2\). Therefore, we have the relationship \(x = \arccos y\). Now we can set up the integral to determine the volume: $$V = \pi \int_{0}^{1} R^2(y) dy$$ where \(R(y)\) is the radius of each washer as a function of \(y\).

Find the radius function

To find the radius function, we observe that the radius is determined by the distance between the curve and the \(y\)-axis, and we already found the relationship \(x = \arccos y\). Therefore, the radius function is: $$R(y) = \arccos y$$ Now, we will substitute this function into the volume integral equation from Step 1.

Evaluate the integral

Now we will substitute our radius function into our integral and evaluate it to determine the volume: $$V = \pi \int_{0}^{1} (\arccos y)^2 dy$$ To evaluate this integral, we will use integration by parts: Let \(u = \arccos y\) and \(dv = u dy\). Then, \(du = -\frac{1}{\sqrt{1-y^2}} dy\) and \(v = y\). $$V = \pi (uv - \int v du)$$ $$V = \pi (\arccos y \cdot y - \int y(-\frac{1}{\sqrt{1-y^2}} dy))$$ The integral now becomes a bit easier to solve: $$V = \pi (\arccos y \cdot y + \int y (\frac{1}{\sqrt{1-y^2}} dy))$$ We recognize this integral as being the result of the inverse substitution (also known as the sin substitution): Let \(y = \sin t\) => \(dy = \cos t dt\). $$V = \pi (\arccos y \cdot y + \int \sin t (\frac{\cos t}{\cos t} dt))$$ So we only need to evaluate: $$V = \pi (\arccos y \cdot y + \int \sin t dt)$$ $$V = \pi (\arccos y \cdot y -\cos t|_{t=0}^{\pi -\arccos y })$$ Now we substitute \(y=\sin t\) and \(1= \sin(\pi - \arccos y)\). $$V = \pi [\arccos (\sin t) \cdot \sin t - \cos t |_{t=0}^{\pi -\arccos(\sin t )}]$$ We evaluate at the bounds and subtract: $$V = \pi [\arccos(\sin(\pi -\arccos(\sin t))) \cdot \sin(\pi -\arccos(\sin t)) - \cos(\pi -\arccos(\sin t))] - \pi [\arccos(\sin(0)) \cdot \sin(0) - \cos(0)]$$ $$V = \pi [0] - \pi [\arccos 0]$$ $$V = -\pi [0 - \pi / 2]$$ $$V = \boxed{\frac{\pi^2}{2}}$$ The volume of the solid generated when the region bounded by \(y=\cos x\) and the \(x\) -axis on the interval \([0, \pi / 2]\) is revolved about the \(y\) -axis is \(\frac{\pi^2}{2}\).