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MAT235PS5-SOLNS - UNIVERSITY OF TORONTO DEPARTMENT OF...

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UNIVERSITY OF TORONTO DEPARTMENT OF MATHEMATICS MAT235Y–CALCULUS II FALL-WINTER 2008-09 ASSIGNMENT 5. DUE ON FEBRUARY 12. PROBLEMS AND SOLUTIONS Problem 1. The rectangular prism pictured below has a rectangular base and a top that is a portion of the plane z = ax + by + c . The four vertical edges can have different lengths. Show by double integration that volume of prism = (area of base) × (average of the lengths of vertical edges) . This formula can be thought of as generalizing the formula for the area of a trapezoid. Solution: Let x 0 and y 0 be as in the following figure Then the volume of the rectangular prism is given by Volume = Z x 0 0 Z y 0 0 ( ax + by + c ) dydx = Z x 0 0 axy + by 2 2 + cy y = y 0 y =0 dx = Z x 0 0 axy 0 + by 2 0 2 + cy 0 dx = ax 2 y 0 2 + bxy 2 0 2 + cxy 0 y = y 0 y =0 = ax 2 0 y 0 2 + bx 0 y 2 0 2 + cx 0 y 0 . 1
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On the other hand the area of base of the prism is x 0 y 0 and the average of the lengths of vertical edges is equal to 1 4 (( ax 0 + c ) + ( ax 0 + by 0 + c ) + ( by 0 + c ) + c ) = ax 0 2 + by 0 2 + c. Therefore, volume of prism = (area of base) × (average of the lengths of vertical edges) . Problem 2. (i) Prove that Z a 0 sin x x dx = Z a 0 1 1 + x 2 dx + Z a 0 sin x x - cos a + x sin a 1 + x 2 e - ax dx. Hint: Apply Fubini’s Theorem to the integral RR [0 ,a ] × [0 ,a ] e - xy sin x dA . (ii) Given that sin x x - cos a + x sin a 1 + x 2 3 for all x and a with x 6 = 0, show that lim a →∞ Z a 0 sin x x - cos a + x sin a 1 + x 2 e - ax dx = 0 . (iii) Use (i) and (ii) to prove that Z 0 sin x x dx = π 2 Solution: (i) By Fubini’s Theorem ZZ [0 ,a ] × [0 ,a ] e - xy sin x dA = Z a 0 Z a 0 e - xy sin x dydx = Z a 0 Z a 0 e - xy sin x dxdy. (1) We have Z a 0 Z a 0 e - xy sin x dydx = - Z a 0 e - xy sin x x y = a y =0 dx = - Z a 0 e - ax sin x x dx + Z a 0 sin x x dx. (2) Now let us compute R a 0 e - xy sin x dx . We will do this applying integration by parts twice.
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