Question

How do you calculate the double integral of `(xcos(x+y))dr` where r is the region: 0 less than or equal to x less then or equal to `(2pi)/6`, 0 less than or equal to y less than or equal to `(2pi)/4`?

Answer 1

Could you clarify your question more? For one thing, you are sort of mixing rectangular and polar coordinates in the way you are stating it.

Answer 2

`int_R intxcos(x+y) dR`

Where `R = {(x, y) : 0 <= x <= pi/3, "and" 0 <= y <= pi/2}`

(I've reduced the fractions.)

We can choose whether to integrate first with respect to `x` or w.r.t. `y`. I'll find:

`int_0^(pi/3) int_0^(pi/2) xcos(x+y) dy dx = int_0^(pi/3) [int_0^(pi/2) xcos(x+y) dy] dx`

I'll do the inner integral, then substitute.
`int_0^(pi/2) xcos(x+y) dy = xsin(x+y)]_(y=0)^(y = pi/2) = xsin(x+ pi/2) - xsin(x)`

Now we need to find

`int_0^(pi/3) [xsin(x+ pi/2) - xsin(x)] dx`

OK, maybe that wasn't easiest. Now I have to do 2 integration by parts, but rather than re-starting, we'll keep on this path.

We could replace `sin(x+ pi/2)` by `cos x`, but many students won't notice that and it's probably not any simpler, so I'll keep on with the problem as I have it above.

First Integral:

`int_0^(pi/3) x sin(x+ pi/2) dx`

Let `u=x` and `dv = sin(x+ pi/2) dx`

so `du = dx` and `v = -cos(x+ pi/2)`

`int_0^(pi/3) x sin(x+ pi/2) dx = -xcos(x + pi/2) + int cos(x + pi/2) dx`

`= -xcos(x + pi/2) + sin(x + pi/2)]_0^(pi/3)`

`= [- (pi)/3 cos((5 pi)/6) + sin((5 pi)/6)] - [0 + sin(pi/2)]`

`= - pi/3 (- sqrt3 / 2) +1/2 - 1 = (pi sqrt3)/6 -1/2 = (pi sqrt 3 -3)/6`

Second Integral:

The second integral:
`int_0^(pi/3) [ - xsin(x)] dx` is almost the same as the first integral.
Using the same method, we'll get

`= - (-xcos(x) + sin(x))]_0^(pi/3) = = xcos(x) - sin(x))]_0^(pi/3)`

Which evaluates to:

`(pi)/3 1 /2 - sqrt3/2 -[0cos (0) - sin(0)] = (pi - 3sqrt3)/6`

Adding the two integrals gives us:

`int_0^(pi/3) [xsin(x+ pi/2) - xsin(x)] dx = (pi sqrt 3 -3)/6+(pi - 3sqrt3)/6 = (pi sqrt3 - 3 sqrt3 + pi -3)/6`