Question

Using polar coordinates, evaluate the integral ∫∫sin(2 2) where r is the region 9≤2 2≤49.

Answer 1

Answer : The integral value is: ∫∫sin(r^2) dr dθ = -π(cos(49) - cos(9))

Given: we have 3 ≤ r ≤ 7.

The integral becomes: ∫∫sin(r^2) dr dθ

with r ranging from 3 to 7 and θ ranging from 0 to 2π.

Now we can evaluate the integral:

∫(0 to 2π) dθ ∫(3 to 7) r sin(r^2) dr

To solve the inner integral, we use substitution. Let u = r^2, so du = 2r dr.

Then: (1/2)∫(9 to 49) sin(u) du = [-(1/2)cos(u)](9 to 49) = -(1/2)[cos(49) - cos(9)]

Now, evaluate the outer integral:

∫(0 to 2π) [-(1/2)(cos(49) - cos(9))] dθ = -[(1/2)(cos(49) - cos(9))][θ](0 to 2π)

= -(1/2)(cos(49) - cos(9))(2π - 0) = -π(cos(49) - cos(9))

So the integral value is:

∫∫sin(r^2) dr dθ = -π(cos(49) - cos(9))