# UT M 408C - HW 15.3-solutions (7 pages)

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## HW 15.3-solutions

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- Pages:
- 7
- School:
- University of Texas at Austin
- Course:
- M 408c - Differential and Integral Calculus

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lewis scl876 HW 15 3 radin 55315 This print out should have 11 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 Consequently I 10 0 points Evaluate the double integral Z Z 4y I dxdy 2 2 D x 1 when D is the region n x y 0 x 1 in the xy plane 1 002 Evaluate the double integral Z Z I x3 y 6xy 2 dA D when o 0 y x D n o x y 0 x 1 x y x 3 5 2 I 2 ln 2 2 I 1 5 3 I 1 3 I 2 5 4 I ln 2 4 4 I correct 5 5 I 2 5 I 1 1 I 4 ln 2 1 6 I correct 2 Explanation As an iterated integral integrating first with respect to y we see that I 0 Now Z 0 1 nZ 0 x 10 0 points 1 I Z 1 2 o 4y dy dx x2 1 2 Explanation The double integral can be rewritten as the repeated integral Z 1 Z x n o I x3 y 6xy 2 dy dx 0 x integrating first with respect to y Now Z x n o x3 y 6xy 2 dy x x 4y dy 2 x 1 2 h i x 2y 2 x 2 x2 1 2 0 x2 1 2 In this case Z 1 I 2 0 h 1 i1 x dx 2 x2 1 2 x 1 0 h1 2 x3 y 2 2xy 3 ix x 4x4 x 2 x2 remember Consequently I 4 003 Z 1 0 x4 dx 10 0 points 4 5 lewis scl876 HW 15 3 radin 55315 over the triangular region A enclosed by the graphs of Evaluate the double integral Z Z I 2x sin y dxdy x 1 D when D is the bounded region enclosed by the graphs of y x2 y 0 2 x y 2 y 4 0 is the surface x 1 1 I 2 sin 1 1 2 I 2 1 cos 1 3 I 1 cos 1 4 I sin 1 1 5 I 1 sin 1 correct Find the volume V of the solid under this graph and over the region A 6 I cos 1 1 1 V 6 ln 6 7 I 2 1 sin 1 2 V 5 ln 6 correct 8 I 2 cos 1 1 Explanation After integration with respect to y we see that Z 1h ix2 2x cos y dx I 0 0 2 Z 1 4 V 6 ln 6 5 V 7 ln 6 Explanation As the region of integration is given by x 1 cos x2 dx 0 3 V 5 ln 6 h x2 sin x2 i1 0 1 1 x 2 x using substitution in the second integral Consequently I 1 sin 1 004 10 0 points The graph of

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