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 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
(a) The integral expression that gives the area of R can be written as:
∫[pi/2, a] (e^x - √(cos(x))) dx
where 'a' is the x-coordinate of the point where the graphs of f(x) = √(cos(x)) and g(x) = e^x intersect.
(b) The volume of the solid generated when R is revolved about the x-axis can be found using the method of cylindrical shells. The integral expression for the volume is:
∫[pi/2, a] 2πx(e^x - √(cos(x))) dx
where 'a' is the x-coordinate of the point where the graphs of f(x) = √(cos(x)) and g(x) = e^x intersect.
(c) The volume of the solid whose cross-sections perpendicular to the x-axis are semicircles with diameters on the XY-plane can be found using the method of cross-sectional areas. The integral expression for the volume is:
∫[pi/2, a] π(√(cos(x)))^2 dx
where 'a' is the x-coordinate of the point where the graphs of f(x) = √(cos(x)) and g(x) = e^x intersect.
(a) To find the area of region R, we need to find the integral expression for the area between the two curves.
The graph of f(x) = sqrt(cos(x)) and the graph of g(x) = e^x intersect at some point(s) in the first quadrant, and the region R is enclosed between these two curves and the vertical line x = π/2.
To find the integral expression, we need to find the x-values at which the two curves intersect.
Setting f(x) = g(x), we have:
sqrt(cos(x)) = e^x
To solve this equation for x, we need to square both sides:
cos(x) = (e^x)^2
cos(x) = e^(2x)
To solve this equation further, we can take the natural logarithm of both sides:
ln(cos(x)) = 2x
Now, we have the expression that gives the x-values at which these two curves intersect.
Next, we need to find the limits of integration for the integral expression for the area of R.
One limit of integration is x = π/2, as mentioned in the problem.
The other limit of integration is the x-value at which f(x) and g(x) intersect.
Let's call this value x = a.
Therefore, the integral expression for the area of region R is:
∫[a, π/2] (g(x) - f(x)) dx
(b) To find the volume of the solid generated when the region R is revolved about the x-axis, we need to use the method of cylindrical shells.
The volume of a cylindrical shell is given by the formula:
V = 2π * ∫[a, b] x * (g(x) - f(x)) dx
In this case, since the horizontal cross-sections are semicircles with diameters on the x-y plane, the radius of each shell is x, and the height is (g(x) - f(x)).
Therefore, the integral expression for the volume of the solid generated is:
V = 2π * ∫[a, π/2] x * (g(x) - f(x)) dx
(c) To find the volume of the solid where the cross-sections perpendicular to the x-axis are semicircles with diameters on the x-y plane, we need to use the method of disks/washers.
The volume of a disk/washer is given by the formula:
V = π * ∫[a, b] (r(x))^2 dx
In this case, since the cross-sections are semicircles with diameters on the x-y plane, the radius of each disk/washer is (g(x) - f(x))/2.
Therefore, the integral expression for the volume of the solid is:
V = π * ∫[a, π/2] ((g(x) - f(x))/2)^2 dx
(a) To find the area of region R, we need to integrate the difference between the graphs of f(x) and g(x) over the interval [0, π/2].
Let A(x) represent the area between the graphs of f(x) and g(x) at a given x-value.
To express the area of R, we have to subtract the upper function (g(x)) from the lower function (f(x)) and integrate over the interval [0, π/2]:
∫[0,π/2] (f(x) - g(x)) dx
(b) To find the volume of the solid generated by revolving region R about the x-axis, we need to use the method of cylindrical shells.
The volume of an infinitesimally thin cylindrical shell with height h, radius r, and thickness dx is given by 2πrh*dx.
In this case, the radius of each cylindrical shell is given by the difference between the upper function (g(x)) and the lower function (f(x)), and the height is determined by the x-coordinate.
To find the total volume, we integrate the volume of each cylindrical shell over the interval [0, π/2]:
V = ∫[0,π/2] 2π[(g(x) - f(x)) * x] dx
(c) To find the volume of the solid when the cross-sections perpendicular to the x-axis are semicircles with diameters on the XY-plane, we can use the method of disks (or washers).
The volume of an infinitesimally thin disk (or washer) with radius r and thickness dx is given by πr^2*dx.
In this case, the radius of each disk is half the distance between the upper function (g(x)) and the lower function (f(x)). The thickness dx of each disk is determined by the x-coordinate.
To find the total volume, we integrate the volume of each disk over the interval [0, π/2]:
V = ∫[0,π/2] π[(g(x) - f(x))/2]^2 dx
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