Question

Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of y=sinx and y=cosx.

Answer 1

since sin and cos = each other at pi/4; take your integrals from 0 to pi/4
[S] cos(t) dt - [S] sin(t) dt ;[0,pi/4]

to revolve it around the x axis;
we do a sum of areas [S] 2pi [f(x)]^2 dx

take the cos first and subtract out the sin next; like cutting a hole out of a donuts.
pi [S] cos(x)^2 dx - [S] sin(x)^2 dx ; [0,pi/4]

cos(2t) = 2cos^2 - 1
cos^2 = (1+cos(2t))/2

1/sqrt(2) - (-1/sqrt(2) +1)
1/sqrt(2) + 1/sqrt(2) -1
(2sqrt(2) - sqrt(2))/sqrt(2) = sqrt(2)/sqrt(2) = 1