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16.3 Double Integrals in Polar Coordinates

In rectangular coordinates, a point is located as (x, y).

Cartesian plot with a point at (3, 4).

Polar Coordinates

Polar: (r, θ)

  • r: displacement from origin
  • θ: angle of line thru origin and the point
Polar plot showing a point at (1, pi/4) with angle and radius labeled.

Conversion

\[x^2 + y^2 = r^2\]

\[x = r \cos \theta\]

\[y = r \sin \theta\]

In Cartesian

\[\iint_R f(x,y) dA\]

dA = area of small patch of region R from change in x and change in y

Diagram of a region R with a differential area element dA = dx dy.

dA = dx dy or dy dx

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In Polar

\[\iint_R f(r, \theta) dA\]

dA = area of a small patch of R from change in r, θ

Polar sector showing a differential area element dA.

Note: this is not a rectangle

Area of this?

Close-up of a polar area element with curved sides r1 dθ and r2 dθ.

When we integrate we sum up infinitely-many of these so dθ is small and dr is small → r1 ≈ r2 = r

Rectangular approximation of the polar area element with dimensions dr and r dθ.

area is dA = r dθ dr = r dr dθ

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Example: Double Integral in Cartesian Coordinates

Evaluate the double integral:

\[ \iint_R \cos(x^2 + y^2) \, dA \]

where \( R \) is the region defined as a circle centered at \( (0,0) \) with radius 2 in the 3rd quadrant.

Region Analysis

The boundary of the circle is given by:

\[ x^2 + y^2 = 4 \]
  • Upper half: \( y = \sqrt{4 - x^2} \)
  • Lower half: \( y = -\sqrt{4 - x^2} \)
Coordinate graph of a circle with radius 2, with the third quadrant region shaded.

Setting up as a Type I Region

The region \( R \) can be expressed in Cartesian coordinates as:

\[ R = \{ (x, y) : -2 \le x \le 0, -\sqrt{4 - x^2} \le y \le 0 \} \]

The integral in Cartesian coordinates is:

\[ \int_{-2}^{0} \int_{-\sqrt{4 - x^2}}^{0} \cos(x^2 + y^2) \, dy \, dx \]
This is hard to evaluate.

Try this in polar coordinates.

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Conversion to Polar Coordinates

In polar coordinates, the region \( R \) (the 3rd quadrant of a circle with radius 2) is defined as:

\[ R = \{ (r, \theta) : 0 \le r \le 2, \pi \le \theta \le \frac{3\pi}{2} \} \]
Polar plot showing the third quadrant region from pi to 3 pi over 2 with radius 2.

Integral Transformation

Recall the polar substitutions:

  • \( x^2 + y^2 \) in polar is \( r^2 \)
  • \( dA \) in polar is \( r \, dr \, d\theta \)
\[ \iint_R \cos(x^2 + y^2) \, dA = \int_{\pi}^{\frac{3\pi}{2}} \int_{0}^{2} \cos(r^2) \, r \, dr \, d\theta \]
\[ = \int_{\pi}^{\frac{3\pi}{2}} \int_{0}^{2} r \cos(r^2) \, dr \, d\theta \]

Evaluation

Using u-substitution for the inner integral:

\[ u = r^2, \quad du = 2r \, dr \]

Continuing the calculation:

\[ = \dots = \frac{\pi}{4} (\sin 4) \]
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Example: Double Integral in Cartesian Coordinates

\[ \int_{0}^{1} \int_{\sqrt{x-x^2}}^{\sqrt{1-x^2}} (x^2 + y^2)^{3/2} \, dy \, dx \]

In Cartesian, this is terrible. Let's examine the region \( R \):

\( 0 \le x \le 1 \)

\( \sqrt{x-x^2} \le y \le \sqrt{1-x^2} \)

Analyzing the Upper Bound

\( y = \sqrt{1-x^2} \) is the upper half circle centered at the origin with radius 1.

Analyzing the Lower Bound

\( y = \sqrt{x-x^2} \)

\( y^2 = x - x^2 \)

\( x^2 - x + y^2 = 0 \)

\( (x - \frac{1}{2})^2 + y^2 = \frac{1}{4} \)

This is the upper half of a circle with radius \( \frac{1}{2} \) centered at \( (\frac{1}{2}, 0) \).

Coordinate graph showing a large blue circle and a smaller green circle. Shaded region R is between them.

Rewrite \( R \) in Polar

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Graph in the first quadrant showing region R bounded by two circular arcs with a radial line segment.

Determining Bounds

Bounds for \( \theta \) is easy: \( 0 \le \theta \le \pi/2 \)

Bounds for \( r \): draw a line from origin to outer edge of \( R \) and see the portion of line inside \( R \).

Small circle \( \le r \le \) big circle

Small Circle: Equation in Polar

\( y^2 = x - x^2 \)

\( x^2 + y^2 = x \)

\( r^2 = r \cos \theta \)

\( r = \cos \theta \)

Big Circle: Equation in Polar

\( r = 1 \)

\( R = \{ (r, \theta), \quad 0 \le \theta \le \pi/2, \quad \cos \theta \le r \le 1 \} \)

Integrate this first because not bounded by constants.

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Integral in Polar Coordinates

The following integral is evaluated by converting to polar coordinates. Note that the term dA is replaced by r dr dθ and the expression x2 + y2 is replaced by r2.

\[ \int_{0}^{\pi/2} \int_{\cos \theta}^{1} (r^2)^{3/2} \, r \, dr \, d\theta \]

Simplifying the integrand:

\[ = \int_{0}^{\pi/2} \int_{\cos \theta}^{1} r^4 \, dr \, d\theta = \dots = \begin{array}{|c|} \hline \frac{\pi}{10} - \frac{8}{75} \\ \hline \end{array} \]
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Example: Volume of a Solid

Example: Find the volume of the solid bounded above by z = 4 - x2 - y2 (a paraboloid) and below by z = 3 (a plane).

Integrate over R which is the "shadow" of the solid on the xy-plane.

3D coordinate system showing a paraboloid opening downwards from z=4, intersected by a plane at z=3.

Finding the Region of Integration

To find the circle radius, find the intersection of z = 3 and z = 4 - x2 - y2:

\[ 3 = 4 - x^2 - y^2 \]\[ x^2 + y^2 = 1 \]
2D xy-plane showing a shaded unit circle labeled R, representing the projection of the solid.

Integration Bounds

  • Bounds: \( 0 \le \theta \le 2\pi \)
  • Radius: \( 0 \le r \le 1 \)
Integrate top (paraboloid) - bottom (plane z=3) over R.
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\[ (4 - x^2 - y^2) - (3) \] \[ = 4 - r^2 - 3 = 1 - r^2 \]
\[ \int_{0}^{2\pi} \int_{0}^{1} (1 - r^2) r \, dr \, d\theta = \dots = \boxed{\frac{\pi}{2}} \]