s11wk10su - Math 23 B. Dodson Week 5a Homework: 15.5...

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Unformatted text preview: Math 23 B. Dodson Week 5a Homework: 15.5 Applications of Double Integrals 15.6 Triple integrals 15.7 Cylindrical Coordinates; 15.8 Spherical Coordinates Problem 15.5.6: Find the mass and center of mass of a thin plate (lamina) occupying the triangular region with verticies at (0, 0), (1, 1) and (4, 0), if the density at (x, y ) is ρ(x, y ) = x. Solution: ρ(x, y ) dA, We have mass m = D and need an iterated integral to evaluate the double integral. Checking the sketch, we see that the region is of Type II, with 0 ≤ y ≤ 1, and y ≤ x ≤ 4 − 3y (where the line from (1, 1) to (4, 0) has slope − 1 , and we solve 3 1 y − 0 = − 3 (x − 4) for x). The iterated integral 4−3y 1 is then x dxdy. Evaluation gives m = 0 10 3. y For the center of mass, (x, y ), we have x = 1 m xρ(x, y ) dA, D 2 and y = 1 m yρ(x, y ) dA. The iterated integrals have D the same limits as above, but in x the integrand is xρ = x2 , and in y the integrand is yρ = xy. The integrals evaluate to 7 and 1, so dividing by mass, (x, y ) = 1 10 3 (7, 1) = (2.1, 0.3). Problem 15.6.x Evaluate the iterated integral z 1 x+z 6xz dydxdz. 0 0 0 Solution. The first inside integral has the anti-derivative 6xz dy = 6xyz, so the definite x+z y =x 6xz dy = [6xyz ]y=0 +z = 6x(x + z ). integral is 0 From this point, the problem is a double iterated integral, with value =1. Finally, we covered problems 15.7.9 (used for #35) then 15.8.10 and 15.8.18. ...
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This note was uploaded on 11/24/2011 for the course MATH 23 taught by Professor Yukich during the Spring '06 term at Lehigh University .

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s11wk10su - Math 23 B. Dodson Week 5a Homework: 15.5...

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