How do I draw the region bounded by the curves, and use the disk method to find the volume when the region is rotated around the x-axis, for y = sin x, y = cos x, and x = 0?

Question

looking for step by step, and with graph please. Thank you.

Answer 1

The light blue is y = cos(x) and the red is y = sin(x). The area bounded by these curves and x = 0 (the y-axis) is shown in yellow. Rotating that area around the x-axis and taking a cross section at the magenta line, results in a "washer" where the blue outside line is the height of the y = cos(x) function and the red inside line is the height of the y = sin(x) function. The area of that cross section can be calculate as:

A = π(r_o² - r_i²) but since r_o = cos(x) and r_i = sin(x) the A = π[cos²(x) - sin²(x)] and since we are finding volume, we would add up all the "washers" from x = 0 to x = π/4 (the place where the two curves intersect) and we would do that by integration:

V = ₀∫^π/4A dx = V = ₀∫^π/4π[cos²(x) - sin²(x)] dx = π₀∫^π/4cos²(x) - sin²(x) dx = π₀∫^π/4cos(2x) dx

The antiderivative of cos(2x) is (1/2)sin(2x) and we evaluate that between zero and π/4:

π[(1/2)sin(2•(π/4) - 0] = π[(1/2)(1) = π/2

Answer 2

Hello.

It is hard to visualize it here. You need to use desmos or graphic utility to see what's bounded region.

1. graph y=sinx & y=cosx
2. You need to be careful here. bounded region is x=0 (y-axis, cosx and sinx)
3. then you will see the region right beside y-axis.
4. x=0 ~ x=pi/4.
5. I will draw rectangle vertically -> its width is dx and if we rotate it, it will be the washer.
6. Bigger circle radius R = cos and smaller circle radius r=sinx
7. pi * integral (0,pi/4) of (R^2 - r^2) dx = pi S (0,pi/4) of (cos^2(x)-sin^2(x)) dx

[cos^2(x) - sin^2(x) = cos(2x)]

= pi * S (0,pi/4) of cos(2x) dx = pi * sin(2x)/2 ] 0,pi/4

=pi * (1/2 * sin (pi/2) - 1/2 * sin0)

=pi * 1/2 = pi/2