Find the volume of the described solid.
The base of a solid is the region between the curve y=6cos(x) and the x-axis from x=0 to x=pi/2. The cross sections perpendicular to the x-axis are squares with diagonals running from the x-axis to the curve.
The answer is 9pi/2 but I don't know how to get this answer...
So the diagonal is of length 6cos(x). If the diagonal of a square is d then the sides of length s follow `s^2+s^2=d^2` so `2s^2=d^2` and since the area of a square is `s^2` , we get the area is `d^2/2 or (6cos(x))^2/2` .
So we add the area of each square with it's thickness (dx), from 0 to `pi/2`
So we integrate (think add) `(6cos(x))^2/2` from 0 to `pi/2` .
`int_0^(pi/2) 36cos^2(x)/2 dx = 18 int_0^(pi/2) cos^2(x) dx `
Since `cos(2x) = 2cos^2(x) - 1 => cos^2(x) = 1/2(cos(2x)+1)` we get
`V = 18 int_0^(pi/2) 1/2(cos(2x) +1)dx = 9 int_0^(pi/2) (cos(2x) + 1) dx`
`V = 9 (1/2sin(2x) + x)|_0^(pi/2) = 9(1/2sin(pi) + pi/2 - (1/2sin(0) + 0)) = (9pi)/2` . since sin(pi) = sin(0) = 0.
So our answer is `V = (9pi)/2`