How Do You Integrate These Challenging Double Integrals?

[hytuoc]

Plz help me integrating the integral below...I did it to a certain point and got stuck...here is the integral and what I did:
1)
Integral form 0 to pi/2, integral from 0 to a*sin(2*theta), [ r ]dr dtheta
Inner integral: Int from 0 to a*sin(2theta) [(r^2)/2] dr = [a^2 * (sin(2 theta))^2 ] / 2
Outer integral: Int from 0 to pi/2 [a^2 * (sin(2 theta))^2 ] / 2] dtheta
...this is when I don't know how to integrate...please show me!

2)
Integral from 0 to pi/2, integral from 0 to 1 [sin(r^2)]dr dtheta...how do I integrate this? (how do I integrate sin(r^2)?)
Thanks

 

[dextercioby]

In the 2) exercise,u're missing "r" from the surface element in polar plane coordinates.That "r" should ease your calculations.

As for the first,write it like that.Denote it by J:
J=a\int_{0}^{\frac{\pi}{2}} \sin 2\theta \ d\theta \int_{0}^{a} r \ dr

And now integrate.

Daniel.

 

[M98Ranger]

1) To continue solving the double integral, we can use the substitution u = sin(2theta) to simplify the inner integral. This will give us:

Inner integral: Int from 0 to a*sin(2theta) [(r^2)/2] dr = [u^2 * a^2] / 4

Now, we can substitute this into the outer integral:

Outer integral: Int from 0 to pi/2 [a^2 * (sin(2 theta))^2 ] / 2] dtheta = Int from 0 to pi/2 [u^2 * a^2] / 4 dtheta

Using the power rule for integration, we can solve this integral to get:

Int from 0 to pi/2 [u^2 * a^2] / 4 dtheta = [a^2 * u^3] / 12 from 0 to pi/2 = [a^2 * (sin(2 theta))^3] / 12 from 0 to pi/2 = [a^2 * (1)^3 - a^2 * (0)^3] / 12 = a^2 / 12

Therefore, the final solution for the double integral is a^2 / 12.

2) To integrate sin(r^2), we can use the substitution u = r^2. This will give us:

Integral from 0 to pi/2, integral from 0 to 1 [sin(r^2)]dr dtheta = Integral from 0 to pi/2, integral from 0 to 1 [sin(u)](1/2u) du dtheta

Using the power rule for integration, we can solve this integral to get:

Integral from 0 to pi/2, integral from 0 to 1 [sin(u)](1/2u) du dtheta = (1/2) Integral from 0 to pi/2, integral from 0 to 1 [sin(u)] du dtheta = (1/2) Integral from 0 to pi/2 [1 - cos(u)] du = (1/2) [u - sin(u)] from 0 to pi/2 = (1/2) [1 - sin(1)] = (1 - sin(1)) / 2

Therefore, the final solution for the double integral is (1 - sin(1)) /