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Evaluating Double Integrals in Polar Coordinates Examples 1
Recall from the Evaluating Double Integrals in Polar Coordinates page that sometimes evaluating a double integral over a region may be difficult due to the nature of the region, and the double integral may be more easily expressible in terms of polar coordinates.
For regions in the form $D = \{ (r, \theta) : a ≤ r ≤ b, \alpha ≤ \theta ≤ \beta \}$ we have that:
(1)For regions in the form $D = \{ (r, \theta) : h_1(\theta) ≤ r ≤ h_2(\theta), \alpha ≤ \theta ≤ \beta \}$ we have that:
(2)We will now look at some examples of evaluating double integrals on regions defined by polar coordinates.
Example 1
Evaluate the double integral $\iint_D x^2 y \: dA$ where $D$ is the top semicircle centered at $(0, 0)$ and with radius $5$.
Our region of integration is $D = \{ (r, \theta) : 0 ≤ r ≤ 5, 0 ≤ \theta ≤ \pi \}$. The region $D$ is sketched below:
Furthermore, evaluating $f$ at $x = r \cos \theta$ and $y = r \sin \theta$ and applying the trigonometric substitution of $\cos ^2 \theta = 1  \sin ^2 \theta$ and we have that:
(3)Thus we have that:
(4)Notice that we can split these integrals into the product of iterated integrals:
(5)Example 2
Evaluate the double integral $\iint_D \sqrt{a^2 x^2  y^2} \: dA$ where $D$ is the circle of radius $a ≥ 0$ centered at the origin, and prove that the volume of a sphere is $V = \frac{4}{3} \pi a^3$.
We first note that $z = \sqrt{a^2 x^2  y^2}$ represents the top hemisphere of the sphere centered at the origin with radius $r$. The region $D$ can be described in polar coordinates as $D = \{ (r, \theta) : 0 ≤ r ≤ a, 0 ≤ \theta ≤ 2\pi \}$. Now:
(6)Now we can evaluate our double integral as follows:
(7)We now need to use substitution to evaluate the inner integral. Let $u = a^2  r^2$. Then $du = 2r \: dr$ and so $ \frac{1}{2} \: du = r \: dr$ and thus:
(8)Putting this into our iterated integrals from earlier and we have that:
(9)Since this double integral represents the volume of half of a sphere, then the volume of a sphere is twice the double integral above and so $V = \frac{4}{3} \pi a^3$.