Question

What is the value of `int_0^(pi/2) int_0^(2acostheta) r sin theta dr d theta`?

Answer 1

`(2a^2)/3`

Explanation:

Assuming

`I = int_0^(pi/2)(int_0^(2acos theta) r dr)sin theta d theta` we have

`I = int_0^(pi/2) 1/2(2acos theta)^2 sin theta d theta`

`I=2a^2int_0^(pi/2)cos^2 theta sin theta d theta = -2/3a^2 (cos^3theta)_0^(pi/2) = (2a^2)/3`

Answer 2

`int_0^(pi/2) \ int_0^(2acos theta) \ rsin theta \ dr \ d theta =2/3a^2`

Explanation:

We want to evaluate:

`int_0^(pi/2) \ int_0^(2acos theta) \ rsin theta \ dr \ d theta`

We evaluate double integrals by starting with the inner integral and treating any other variables except the variable of integration as constant.

Here the inner integral (treating `theta` as a constant) is evaluated as follows:

`int_0^(2acos theta) \ rsin theta \ dr = sin theta \ int_0^(2acos theta) \ r \ dr`
`" " = sin theta \ [1/2r^2]_0^(2acos theta)`

`" " = 1/2sin theta \ [r^2]_0^(2acos theta)`

`" " = 1/2sin theta \ ((2acos theta)^2-0)`

`" " = 1/2sin theta \ (4a^2cos^2 theta)`

`" " = 2a^2 \ sin theta \ cos^2 theta`

And so we can now evaluate the double integral as follows:

`int_0^(pi/2) \ int_0^(2acos theta) \ rsin theta \ dr \ d theta = int_0^(pi/2) \ 2a^2 \ sin theta \ cos^2 theta \ d theta`
`" " = -2/3a^2 \ int_0^(pi/2) \ 3cos^2 theta (-sin theta) \ d theta`
`" " = -2/3a^2 \ [cos^3 theta]_0^(pi/2)`

`" " = -2/3a^2 \ (cos^3(pi/2) - cos^3(0))`

`" " = -2/3a^2 \ (0-1)`

`" " = 2/3a^2`