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week09 - the surface z = xy that lies within the cylinder x...

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Math 23 Sections 110-113 B. Dodson Week 9 Homework: 15.5 Applications (mass, center of mass) 15.6 Surface Area 15.7 triple integrals 15.8 cylindrical, spherical coords 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: We have mass m = ZZ D ρ ( x, y ) dA, 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 - 3 y (where the line from (1 , 1) to (4 , 0) has slope - 1 3 , and we solve y - 0 = - 1 3 ( x - 4) for x ). The iterated integral is then Z 1 0 Z 4 - y y x dxdy. Evaluation gives m = 10 3 . For the center of mass, ( x, y ) , we have x = 1 m ZZ D ( x, y ) dA,

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2 and y = 1 m ZZ D ( x, y ) dA. The iterated integrals have the same limits as above, but in x the integrand is = x 2 , and in y the integrand is = xy. The integrals evaluate to 7 and 1, so dividing by mass, ( x, y ) = 1 10 3 (7 , 1) = (2 . 1 , 0 . 3) . Problem 15.6.9. Find the surface area of the part of
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Unformatted text preview: the surface z = xy that lies within the cylinder x 2 + y 2 = 1 . Solution. With f ( x, y ) = xy, the surface area formula has f x = y, f y = x, so p y 2 + x 2 + 1 and Area(S) = ZZ D p y 2 + x 2 + 1 dA, where D is the region inside the circle x 2 + y 2 = 1 . Switching to polar coords, we get the iterated integral Z 2 π Z 1 p r 2 + 1 r drdθ. Problem 15.7.3. Evaluate the iterated integral Z 1 Z z Z x + z 6 xz dydxdz. Solution. The ﬁrst inside integral has the anti-derivative Z 6 xz dy = 6 xyz, so the deﬁnite integral is Z x + z 6 xz dy = [6 xyz ] y = x + z y =0 = 6 x ( x + z ) . From this point, the problem is a double iterated integral, with value =1. 3 Finally, we covered problems 15.7.9 (used for #35) then 15.8.10 and 15.8.18 all of which are past the syllabus for Exam 2, but will be needed in Ch. 16....
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