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Double Integrals in Polar Coordinates

15.4 Double Integrals in Polar Coordinates

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Double Integrals in Polar Coordinates

Suppose that we want to evaluate a double integral: R f (x, y) dA, where R is one of the regions shown below. In either case the description of R in terms of rectangular coordinates is rather complicated, but R is easily described using polar coordinates.

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Double Integrals in Polar Coordinates

Polar coordinates (r, ) of a point are related to the rectangular coordinates (x, y) by the equations:

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Double Integrals in Polar Coordinates

Therefore we have: Be careful not to forget the additional factor r on the right side! A method for remembering this: Think of the “infinitesimal” polar rectangle as an ordinary rectangle with dimensions r d and dr and therefore has “area” dA = r dr d.

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Example 1 Evaluate R (3x + 4y2) dA, where R is the region in the upper half-plane bounded by the circles x2 + y2 = 1 and x2 + y2 = 4. Solution: The region R can be described as: R = {(x, y) | y 0, 1 x2 + y2 4} It is a half-ring and in polar coordinates it is given by: 1 r 2, 0 .

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Example 1 – Solution cont’d Rewrite in terms of polar coordinates:

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Double Integrals in Polar Coordinates

What we have done so far can be extended to the more complicated type of region In fact, by combining Formula 2 with where D is a type II region, we obtain the following formula: Figure 7

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Practice 1: Answer:

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Integrate in r: Integrate in q:

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Practice 2: Where D is the bottom half of : Answer:

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Integrate in r: Integrate in q:

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Practice 3: Answer: Region is defined by:

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So the integral is: And we integrate in r then in q:

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Applications of Double Integrals

15.5

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2. Charge Density and Charge

1. Density and Mass 2. Charge Density and Charge

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2. Charge density and total Charge:

1. Density and Mass: Total mass m of an object with variable density (x, y): 2. Charge density and total Charge: Total charge Q of a region D with charge density (x, y) :

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Example Charge is distributed over the triangular region D so that the charge density at (x, y) is (x, y) = xy, measured in coulombs per square meter (C/m2). Find the total charge.

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Example – Solution From Equation 2 and the graph of the region, we can write: The total charge is C.

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Another application of double integrals:

15.6 Surface Area

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Surface Area Let S be a surface with equation z = f (x, y), where f has continuous partial derivatives:

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Surface Area If we use the alternative notation for partial derivatives, we can rewrite Formula 2 as follows: Notice the similarity between the surface area formula in Equation 3 and the arc length formula in calc 2:

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Example Find the surface area of the part of the surface z = x2 + 2y that lies above the triangular region T in the xy-plane with vertices: (0, 0), (1, 0), and (1, 1). Solution: The region T is shown below and is described by: T = {(x, y) | 0 x 1, 0 y x}

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Example – Solution Using Formula 2 with f (x, y) = x2 + 2y, we get:

cont’d Using Formula 2 with f (x, y) = x2 + 2y, we get: Here is the graph of the area portion we have just computed:

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