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

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## HW 15.3-solutions

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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 f x y 1 x y 4 6 4 1 4 x 4 not drawn to scale the double integral can be written as the repeated integral Z 6 Z 2 x 1 I dy dx x y 4 1 4 lewis scl876 HW 15 3 radin 55315 integrating first with respect to y from y 4 to y 2 x Now the inner integral is equal to h i2 x ln 6 ln x ln x y 4 4 Thus I Z 1 6n o ln 6 ln x dx h i6 5 ln 6 x ln x x 1 Consequently V 5 ln 6 005 10 0 points 3 x and y x2 the volume of the solid is given by the double integral V Z 0 1Z x 2x 3y dydx x2 since the graphs of y 2 x and y x2 intersect at 0 0 and 1 1 But then after integration with respect to y Z 1h 3 2i x V 2xy y dx 2 x2 0 Z 1n o 3 4 3 2 3 2 x x x x dx 2 0 h 2 1 3 1 2 1 5 i1 2 x5 2 x4 x x 5 4 2 2 5 0 Consequently Determine the volume of the solid lying under the graph of I 9 3 3 10 20 4 006 10 0 points z 2x 3y and above the bounded region enclosed by the graphs of y 2 x and y x2 1 volume 11 20 2 volume 13 20 3 volume 17 20 Find the volume of the solid in the first octant bounded by the cylinders x2 y 2 9 y2 z2 9 Hint in the first octant the cylinders are shown in z 4 volume 3 correct 4 5 volume 9 20 Explanation The volume of the solid lying under the graph of z f x y and above the region A in the xy plane is given by the double integral Z Z V f x y dxdy A Thus when f x y 2x 3y and A is the bounded region enclosed by the graphs of y 2 y 3 3 x lewis scl876 HW 15 3 radin 55315 4 Consequently 1 volume 20 cu units V 2 volume 22 cu units 1 9y y 3 3 3 volume 18 cu units correct 007 D Explanation As the figure shows the solid in the first octant bounded by the cylinders y 3x 12 1 I 13 2 I 14 x2 y 2 9 3 I 12 correct lying in the first quadrant of the xy plane Thus the volume of the solid is given by the double integral Z Z p 9 y 2 dx dy V A where A is the region in the first quadrant of the x y plane bounded by the quarter circle n o p x y 0 x 9 y 2 0 y 3 4 I 15 5 I 16 Explanation The region of integration D is the region shaded in y 3x 12 4 and so V can be represented as the iterated integral Z 3 Z 9 y 2 p 9 y 2 dx dy V In this case 0 12 0 3 h p x 9 y2 Z 3 i 9 y 2 0 2 0 x 12 as well as the x and y axes above that part of the circle V 3 dA 4 x y 3 2 when D is the region in the first quadrant bounded by y2 z2 9 is the solid below the graph of p z 9 y2 Z 10 0 points Z Z I 5 volume 21 cu units 0 18 cu units 0 Evaluate the double integral 4 volume 19 cu units x2 y 2 9 3 9 y dy dy the vertical line interior to the region showing that we should integrate first with respect to y Then I becomes the repeated integral I Z 0 12 Z 3x 12 0 3 dy dx 4 x y 3 2 lewis scl876 HW 15 3 radin 55315 Now the inner integral is h 1 i3x 12 6 3 …

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