Question
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.
(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.
(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.
Answers
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
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
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
∫[0,π/2] 1/2 π(1/2 (e^x - √cosx))^2 dx
= 1/4 ∫[0,π/2] π(e^x - √cosx)^2 dx
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