Let R be the region bounded by the y-axis and the curves y = sin x and y = cos x. Answer the following.
a)Find the exact area of R.
b)A solid is generated by revolving R about the x-axis. Find the exact volume of the solid.
The curves intersect at (pi/4,1/√2)
On that interval, cos(x) > sin(x), so
A = ∫[0,pi/4](cosx-sinx)dx
= sinx+cosx[0,pi/4]
= (1/√2+1/√2)-(0+1) = √2-1
V = ∫[0,pi/4]pi*(R^2-r^2)dx
where R=cosx and r=sinx
V = pi*∫[0,pi/4](cos^2-sin^2)dx
= pi*∫[0,pi/4]cos(2x) dx
= pi/2 sin(2x)[0,pi/4]
= pi/2 * (1-0)
= pi/2
a) To find the exact area of the region R, we need to find the points of intersection between the curves y = sin x and y = cos x. These points will determine the limits of integration to calculate the area.
Setting sin x = cos x, we can rearrange the equation to get tan x = 1. Taking the inverse tangent of both sides, we find x = π/4 as the first solution. To find the second solution, we add π to the result, giving x = 5π/4.
So, the region R is bounded by x = π/4 and x = 5π/4. The area can be calculated using the definite integral of the absolute difference between the curves:
Area = ∫[π/4, 5π/4] |sin x - cos x| dx
To solve this integral, we can split it into two separate integrals:
Area = ∫[π/4, 5π/4] (sin x - cos x) dx + ∫[π/4, 5π/4] (cos x - sin x) dx
Evaluating these integrals gives:
Area = [−cos x - sin x] + [sin x - cos x] evaluated from π/4 to 5π/4
= [−cos(5π/4) - sin(5π/4) + cos(π/4) + sin(π/4)]
Using the values of cos(π/4) = sqrt(2)/2 and sin(π/4) = sqrt(2)/2, we get:
Area = [−(-sqrt(2)/2) - (-sqrt(2)/2) + sqrt(2)/2 + sqrt(2)/2]
= sqrt(2)
Therefore, the exact area of region R is sqrt(2).
b) To find the exact volume of the solid generated by revolving R about the x-axis, we can use the method of cylindrical shells. The volume of each shell can be calculated by multiplying the circumference of the shell by its height and thickness.
The circumference of each shell can be approximated as the distance from the x-axis to the curve y = cos x:
Circumference = 2πy
The height of each shell is the difference between the curves y = sin x and y = cos x:
Height = sin x - cos x
Integrating the product of the circumference and height over the region R gives the volume:
Volume = ∫[π/4, 5π/4] [(2πy)(sin x - cos x)] dx
Since y = cos x, we can substitute this into the equation:
Volume = ∫[π/4, 5π/4] [(2πcos x)(sin x - cos x)] dx
Evaluating this integral gives:
Volume = [−πcos^2 x/2 + πsin^2 x/2] evaluated from π/4 to 5π/4
= [−π/2 + π/2] - [0 - 0]
= π
Therefore, the exact volume of the solid generated by revolving R about the x-axis is π.