91Ó°ÊÓ

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 \(y\).

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.

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 \(\frac{\pi^2}{2}\).