Let R be the region in the first quadrant enclosed by the graph of f(x) = sqrt cosx, the graph of g(x) = e^x, and the vertical line pi/2, as shown in the figure above.

Question

(a) Write. but do not evaluate, an integral expression that gives the area of R.
(b) Find the volume of the solid generated when R is revolved about the x-axis.
(c) Region R is the base of a solid whose cross-sections perpendicular to the x-axis are semicircles with diameters on the XY-plane. Write, but do not evaluate, an integral expression that gives the volume of this solid.

Answer 1

A quick sketch will show that y = √cosx and y = e^x intersect at (0,1) and some negative value of x, but we are to stay in quad I
Also y = √cosx = 0 when x = π/2

So Area needed = (∫ e^x dx - ∫ √cosx dx) from 0 to π/2

volume for your region
= π (∫ (e^x)^2 dx - ∫ cosx dx) from 0 to π/2

Answer 2

Each semicircle has a diameter e^x - √cosx. So the volume is
∫[0,π/2] 1/2 π(1/2 (e^x - √cosx))^2 dx
= 1/4 ∫[0,π/2] π(e^x - √cosx)^2 dx