Question

evaluate the double integral: xcos(y)DA, D is bounded by y=0, y=x^2, x=1

Answer 1

you are integrating x from 0 to 1, but y from 0 to x^2. You are integrating below the parabola, not a square.


int (0 to 1) [int (0 to x^2) x cosy dy] dx =

int (0 to 1) [int (0 to x^2) cos y dy] x dx=

int (0 to 1) [sin y (0 to x^2)] x dx =

int (0 to 1) [sin (x^2) - sin 0 ] x dx =

int (0 to 1) sin (x^2) x dx

(1/2) int (0 to 1) sin (x^2) 2xdx =

1/2 int (0 to 1) sin u du = - (1/2) cos u (0 to 1)

u = x^2, du = 2xdx

u(1) = 1
u(0) = 0

So, the result is: (cos 0 - cos 1)/2