Question
Consider the region R in the first quadrant under the graph of y = cos x from x = 0 to \(x=\frac{\pi }{2}\).
(A) Find the area of R.
(B) What is the volume of the solid obtained by rotating R about the x-axis?
(C) Suppose R is the surface of a concrete slab. If the depth of the concrete at x, where x is given in feet, is \(d(x)=\sin x+1\) find the volume (in cubic feet) of the concrete slab.
Answer/Explanation
(A) The area bounded by the curves is given by
\(\int_{0}^{\pi /2}\cos xdx=[\sin x]^{\pi /2}_{0}=1\)
(B)If you rotate the region about the x-axis, note that each cross section perpendicular to the x-axis is a disc with radius y = cos x. Therefore, you set up the following integral to compute the volume:
\(\pi \int_{0}^{\pi /2}\cos^{2}xdx=\frac{\pi }{2}\int_{0}^{\pi /2}(1+\cos (2x))dx=\frac{\pi }{2}\left [ x+\frac{\sin (2x)}{2} \right ]^{\pi /2}_{0}=\frac{\pi ^{2}}{4} \)
(C)Cross sections perpendicular to the x-axis are rectangles of depth sinx + 1 and width cos x. Therefore, the volume is given by
\(\int_{0}^{\pi /2}(\sin x+1)\cos xdx\)
This integral is computable by many techniques. You will make the substitution u=sin x+1,so that du = cosxdx, and the integral becomes
\(\int_{1}^{2}udu=\frac{1}{2}[4-1]=\frac{3}{2}\) or \(1.5ft^{3}\)