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Unformatted text preview: Math 23 B. Dodson Week 4b Homework: 15.2 iterated integrals 15.3 general regions 15.4 polar coords Week 5a Homework: 15.5 Applications of Double Integrals [Omit #23] 15.6 Triple integrals 15.7 Cylindrical Coordinates; 15.8 Spherical Coordinates Problem 15.2.3: Evaluate the iterated integral integraldisplay 3 1 integraldisplay 1 (1 + 4 xy ) dxdy. Solution: We start with the inside integral: integraldisplay 1 (1 + 4 xy ) dx and find an antiderivative (under ∂ ∂x ) for 1 + 4 xy. Holding y constant, we find ∂ ∂x ( x + 2 x 2 y ) = 1 + 4 xy, so integraldisplay 1 (1 + 4 xy ) dx = bracketleftbig x + 2 x 2 y bracketrightbig 1 = (1 + 2 y ) − (0 + 0) = 1 + 2 y. We replace the inside integral with this value, 2 giving the outside integral: integraldisplay 3 1 (1 + 2 y ) dy = bracketleftbig y + y 2 bracketrightbig 3 1 = (3 + 9) − (1 + 1) = 10 . From 15.1 we covered 15.1.1a. The above iterated integral calculates the value of the double integral integraldisplayintegraldisplay R (1 + 4 xy...
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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 .
 Spring '06
 YUKICH
 Calculus, Integrals

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