Question

11. \(y = co{s^2}x,|x|\pi /2,y = \frac{1}{4};\quad \) about \(x = \pi /2\)

Short answer

\(\int_0^1 \pi \left( {{{\left( {2 - {x^2}} \right)}^2} - {{(2 - \sqrt x )}^2}} \right)\)

Step-by-step solution

To Find the x-axis equation

\(y = co{s^2}x,y = \frac{1}{4},|x| \le \frac{\pi }{2},1\quad \)about \(x = \frac{\pi }{2}\)

The\(|x| \le \frac{\pi }{2}\)part appears to be there so we only use the one enclosed region where\( - \pi /2 \le x \le \pi /2\), otherwise there would be a infinite number of enclosed regions between\(y = co{s^2}x,y = \frac{1}{4}\)

Find the intersections

\(co{s^2}x = \frac{1}{4}\)

\(x = \pm \frac{\pi }{3}\quad \left( { no \pm \frac{{2\pi }}{3} since |x| \le \frac{\pi }{2}} \right)\)

Symmetry can't be used since one half of the symmetric region is further away from the axis of rotation than the other half. So we integrate from \( - \frac{\pi }{3}\) to \(\frac{\pi }{3}\)

To find the cylinder method

Subtract the lower from the upper to get the vertical distances which will be the heights of the cylinders. (subtract a lesser value from a greater value to get positive values)

\(h = co{s^2}x - \frac{1}{4}\)

The radius of each cylinder as\(x\)goes from\( - \frac{\pi }{3}\)to\(\frac{\pi }{3}\)will be

\(r = \frac{\pi }{2} - x\)

which gives the horizontal distance. (subtract a lesser value from a greater, left from right)

Final Answer

The integral with the cylinder method:

\(\begin{aligned}{}V &= 2\pi \int_a^b {( radius )} ( height )dx\\ &= 2\pi \int_{\pi /3}^{\pi /3} {\left( {\frac{\pi }{2} - x} \right)} \left( {co{s^2}x - \frac{1}{4}} \right)dx\end{aligned}\)

\(V = 2\pi \int_{\pi /3}^{\pi /3} {\left( {\frac{\pi }{2} - x} \right)} \left( {co{s^2}x - \frac{1}{4}} \right)dx\)