Question

Sketch the region of integration and convert the polar integral to a Cartesian integral or sum of integrals. Do not evaluate the integral.
∫ ^π∫^2 r^3 sinθcosθdrd()
π/2 0

Answer 1

It looks like the integral in polar coordinates is given to be

$\displaystyle\int_{\pi/2}^\pi \int_0^2 r^3\sin(\theta)\cos(\theta)\,\mathrm dr\,\mathrm d\theta$

Converting back to Cartesian, we take

x = r cos(θ)

y = r sin(θ)

dx dy = r dr dθ

so we can easily recover the integrand in Cartesian:

$r^3\sin(\theta)\cos(\theta)\,\mathrm dr\,\mathrm d\theta = (r\sin(\theta))(r\cos(\theta))(r\,\mathrm dr\,\mathrm d\theta) = xy\,\mathrm dx\,\mathrm dy$

This leaves us with the limits:

• π/2 ≤ θ ≤ π corresponds to the second quadrant of the (x, y)-plane (that is, where x < 0 and y > 0)

• 0 ≤ r ≤ 2 correspond to the disk of radius 2 centered at the origin

Taken together, we see the region of integration is a quarter-disk of radius 2 in the second quadrant, which we can capture by the set

{(x, y) : -√2 ≤ x ≤ 0 and 0 ≤ y ≤ √(2 - x ²)}

So, in Cartesian coordinates, the integral would be

$\displaystyle \boxed{\int_{-\sqrt2}^0 \int_0^{\sqrt{2-x^2}} xy\,\mathrm dy\,\mathrm dx}$