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Dated: 30-10-2024

Double Integral1 In Polar Coordinates

There are two reasons why double integrals1 are important:

  1. They arise naturally in many applications.
  2. They are more easily evaluated in polar coordinates as compared to rectangular coordinates.

Integrals2 In Polar Coordinates

Imagine a region bounded by:

  • \(\theta_1 = \alpha\)
  • \(\theta_2 = \beta\)
  • \(r = r_1 \cdot \theta\)
  • \(r = r_2 \cdot \theta\)

Where \(0 \le r \cdot \theta \le a\) and \(\alpha \le \theta \le \beta\).
Then the double integral1 is given by

\[\iint\limits_{R} f(r, \theta) \, dA = \int_{\theta=\alpha}^{\theta=\beta} \int_{r=r_1(\theta)}^{r=r_2(\theta)} f(r, \theta) \, rdr d\theta\]

Finding Limits of Integration from Sketch

  1. Since \(\theta\) is held fixed for first integral,2 draw a radial line3 at angle \(\theta\). This line crosses the region \(R\) at 2 points which are the boundaries for first integral.2
  2. Rotate the radial line3 such that it intersects the region \(R\). The smallest \(\theta\) is \(\alpha\) and largest is \(\beta\) where we get the intersection with \(R\). Hence, the limits for \(\theta\) are \(\alpha\) and \(\beta\).

Example

Evaluate

\[\iint_R \sin(\theta) dA\]

Where \(R\) is the region in the first quadrant4 that is outside the circle \(r = 2\) and inside the cardioid \(r = 2(1 + \cos (\theta))\).

Solution

Because it is in the first quadrant,4 the limits of \(\theta\) are \(0\) and \(\frac \pi 2\).
For \(r\), the limits are \(1\) and \(2(1 + \cos (\theta))\).

\[\iint_R \sin(\theta) dA = \int_0 ^ {\frac \pi 2}\int_2^{2(1 + \cos(\theta))} \sin (\theta) r dr d\theta\]
\[= \int_0^{\frac \pi 2} \left[\frac 1 2 r^2 \sin(\theta) \right]_{r = 2}^{r = 2(1 + \cos (\theta))} d\theta\]
\[= 2 \int_0 ^ {\frac \pi 2} \left((1 + \cos(\theta))^2 \cdot \sin(\theta) - \sin(\theta) \right) d\theta\]
\[ = 2 \left[ -\frac{1}{3} (1+\cos\theta)^3 + \cos\theta \right]_{0}^{\pi/2} \]
\[ = \left( -\frac{1}{3} - \left(-\frac{5}{3} \right) \right) \]
\[ = \frac{8}{3} \]

Changing Cartesian Integral2 into Polar Integral2

  1. Substitute \(x = r \cdot \cos(\theta)\), \(y = r \cdot \sin(\theta)\) and \(dx dy = r dr d\theta\).
  2. Supply polar limits for the integrals.2
\[\iint_R f(x, y) dx dy = \iint_G f(r \cdot \cos(\theta), r \cdot \sin(\theta)) r dr d\theta\]

Example

Evaluate the following by changing into polar coordinates.

\[\int_0^1 \int_0^{\sqrt{1 - x^2}} (x^2 + y^2) dy dx\]

Solution

From inspection, it is obvious that

\[0 \le x \le 1\]
\[0 \le y \le \sqrt{1 - x^2}\]

The limits tell us that we are interested in the first quadrant4 only
\(y = \sqrt{1 - x^2}\) is a circle.

\[y^2 = 1 - x^2\]
\[x^2 + y^2 = 1\]

This tells us that \(r = 1\)

\[\because x^2 + y^2 = r^2\]
\[= \int_0^{\frac \pi 2}\int_0^1 r^3 dr d\theta\]
\[= \int_0^{\frac \pi 2} \left[\frac {r^4} 4\right]_0^1 d\theta\]
\[= \int_0^{\frac \pi 2} \frac 1 4 d\theta\]
\[= \left[\frac \theta 4\right]_0^{\frac \pi 2}\]
\[\frac \pi 8\]

References

Read more about notations and symbols.

  1. Read more about double integrals. ↩↩↩

  2. Read more about integrals. ↩↩↩↩↩↩

  3. Read more about lines. ↩↩

  4. Read more about quadrants. ↩↩↩