Question:
Let R be the region enclosed by the graphs of \(g(x)= -2 + 3 cos \left ( \frac{\pi }{2}x \right )\) and h(x) = 6 – 2(x-1)2, the y-axis, and the vertical line x = 2, as shown in the figure above.
(a) Find the area of R.
(b) Region R is the base of a solid. For the solid, at each x the cross section perpendicular to the x-axis has area \(A(x)=\frac{1}{x+3}.\) Find the volume of the solid.
(c) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y = 6.
Answer/Explanation
Ans:
(a)
\(Area = \int_{0}^{2}h(x)-g(x)dx\)
\( = \int_{0}^{2}6-2(x-1)^{2}dx – \int_{0}^{2}-2+3 cos (\frac{\pi }{2}x)dx\)
\(=6x – \frac{2}{3}(x-1)^{3}|_{0}^{2}-\left ( -2x+\frac{6}{\pi }sin(\frac{\pi }{2}x) \right )|_{0}^{2}\)
\(=\left ( \left [ 12-\frac{2}{3}(1) \right ]-\left [ 0+\frac{2}{3} \right ] \right )-\left ( \left [ -4+\frac{6}{\pi }(0) \right ]-\left [ 0+0 \right ] \right )\)
\(=12-\frac{4}{3}+4=16-\frac{4}{3}\)
(b)
\(v = \int_{0}^{2}A(x)2x=\int_{0}^{2}\frac{1}{x+3}dx=ln|x+3||_{0}^{2}=ln |5| – ln |3|\)
= ln 5- ln 3
(c)
\(v = \int_{0}^{2}\pi \left [ (4-g(x))^{2}-(6-h(x))^{2} \right ]dx\)