10.28 - disk we have D | x | dA = L | x | dA R | x | dA But...

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Section 12.3: Double integrals in polar coordinates (concluded) Group Work 2, problem 2: I get (13/3) π 3 even though the Instructor’s Guide says the answer is (4/3) 3 . Collect HW and section-notes; return HW Ask students to remind me to stop at 12:40 to discuss exam Section 12.4: Applications of double integrals Main ideas? ..?. . Density and mass; charge density and charge Moment about an axis; centroid Moment of intertia about an axis or about a point Radius of gyration Problem for class discussion: What is the total mass M of a disk of radius 1 if its density is ρ ( x , y ) = | x | + | y |? ..?. . M = ∫∫ D | x |+| y | dA = ∫∫ D | x | dA + ∫∫ D | y | dA . Since D is symmetric in x and y , we have ∫∫ D | x | dA = ∫∫ D | y | dA .

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Hence M = 2 ∫∫ D | x | dA . If we let L and R be the left and right halves of the unit
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Unformatted text preview: disk, we have D | x | dA = L | x | dA + R | x | dA . But we also have L | x | dA = R | x | dA (if this isnt obvious to you, think about Riemann sums). So M = 4 R | x | dA = 4 R x dA . Lets do this two ways: Cartesian coordinates: R x dA = ..?. . 1 sqrt(1 xx ) sqrt(1 xx ) x dy dx = 1 2 x sqrt(1 x 2 ) dx = (2/3)(1 x 2 ) 3/2 | 1 = 2/3 . So M = 4(2/3) = 8/3. Polar coordinates: R x dA = ..?. . /2 /2 1 ( r cos ) r dr d = ..?. . ( /2 /2 cos d ) ( 1 r 2 dr ) = (sin | /2 /2 ) ( r 3 /3 | 1 ) = (2)(1/3) = 2/3 as before. Proceed as above....
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10.28 - disk we have D | x | dA = L | x | dA R | x | dA But...

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