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

a) Find the area of R

b) Find the volume of the solid generated when R is revolved about the x-axis

c) Find the volume of the solid whose base is R and whose cross sections cut by planes perpendicular to the x-axis are squares ... ?

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