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.
Over what interval? The curves intersect in many places.
Sure! Let's start with part (a) and find the exact area of region R.
To find the area of region R, we need to find the definite integral of the difference between the upper and lower curves with respect to x. In this case, the upper curve is y = sin(x) and the lower curve is y = cos(x).
To determine the boundaries of the integral, we need to find the x-values at the points where the two curves intersect. Since both sin(x) and cos(x) have a period of 2π, we can solve the equation sin(x) = cos(x) for x within the interval [0, 2π].
Setting sin(x) = cos(x), we have:
sin(x) - cos(x) = 0
Rearranging the equation, we get:
sin(x)/cos(x) - 1 = 0
Using the identity sin(x)/cos(x) = tan(x), the equation becomes:
tan(x) - 1 = 0
Solving for x, we find:
x = π/4
So, the boundaries of the integral are x = 0 and x = π/4.
Now, the area of region R can be found by evaluating the definite integral:
A = ∫[0,π/4] (sin(x) - cos(x)) dx
Using the integral properties, we can break down this integral into two separate integrals:
A = ∫[0,π/4] sin(x) dx - ∫[0,π/4] cos(x) dx
Integrating each term individually, we have:
A = [-cos(x)] [0,π/4] - [sin(x)][0,π/4]
Evaluating the integral at the boundaries, we get:
A = [-cos(π/4)] - (-cos(0)) - [sin(π/4)] - (-sin(0))
Simplifying, we find:
A = -cos(π/4) + cos(0) - sin(π/4) + sin(0)
Using the values cos(0) = 1 and sin(0) = 0, the equation simplifies further:
A = -cos(π/4) + 1 - sin(π/4)
Finally, substituting the values of cos(π/4) = sin(π/4) = √2/2, we get:
A = -√2/2 + 1 - √2/2
Simplifying, we find:
A = 1 - √2
So, the exact area of region R is 1 - √2.
Now, let's move on to part (b) and find the exact volume of the solid generated by revolving region R about the x-axis.
To find the volume, we need to set up an integral that represents the sum of infinitesimally small volumes of disks with radius y and thickness dx, as we revolve region R around the x-axis.
The infinitesimal volume of each disk is given by:
dV = πy^2 dx
Integrating this equation over the same interval [0, π/4] that we used for the area, we have:
V = ∫[0,π/4] πy^2 dx
Substituting the equations for y = sin(x) and y = cos(x), we get:
V = ∫[0,π/4] π[sin(x)]^2 dx - ∫[0,π/4] π[cos(x)]^2 dx
Evaluating each integral separately, we find:
V = π∫[0,π/4] sin^2(x) dx - π∫[0,π/4] cos^2(x) dx
Using the double angle identity sin^2(x) = (1 - cos(2x))/2 and cos^2(x) = (1 + cos(2x))/2, the equation becomes:
V = π/2 ∫[0,π/4] (1 - cos(2x))/2 dx - π/2 ∫[0,π/4] (1 + cos(2x))/2 dx
Simplifying, we get:
V = π/2 ∫[0,π/4] (1 - cos(2x))/2 - (1 + cos(2x))/2 dx
Combining like terms, we have:
V = π/4 ∫[0,π/4] (1 - cos(2x) - 1 - cos(2x)) dx
Simplifying further, we get:
V = π/4 ∫[0,π/4] -2cos(2x) dx
Using the double angle identity cos(2x) = 1 - 2sin^2(x), the equation becomes:
V = -π/2 ∫[0,π/4] 2(1 - 2sin^2(x)) dx
Expanding and integrating, we get:
V = -π/2 ∫[0,π/4] (2 - 4sin^2(x)) dx
Simplifying and evaluating the integral, we find:
V = -π/2 [2x - 4/3 sin^3(x)] [0,π/4]
Plugging in the boundaries, we get:
V = -π/2 [(2π/4) - 4/3 sin^3(π/4)] - (-π/2 [0 - 4/3 sin^3(0)])
Simplifying, we find:
V = -π/2 [π/2 - 4/3 (√2/2)^3] - (-π/2 [0 - 4/3 (0)])
V = -π/2 [π/2 - 4/3 (1/2√2)] - (-π/2 [0])
V = π/2 [4/3 (1/2√2) - (π/2)]
Simplifying further, we get:
V = π/2 [(2/3)√2 - (π/2)]
So, the exact volume of the solid generated by revolving region R about the x-axis is (π/2)√2 - (π^2)/4.