2220hw6sol

2220hw6sol - Math 2220 Section 5.3 Problem Set 6 Solutions...

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Unformatted text preview: Math 2220 Section 5.3 : Problem Set 6 Solutions Spring 2010 3 ex 6. Sketch the region of integration for the integral 0 1 2 dy dx , then reverse the order of integration and evaluate both integrals, the original one and the one with the order reversed. Solution. The iterated integral is equal to a double integral over the region bounded by the horizontal line y = 1 , the vertical line x = 3 and the curve y = ex . The iterated integral is equal to 3 0 1 ex 3 2 dy dx = 0 2y ex 1 3 dx = 0 2(ex − 1) dx = (2ex − 2x) = 2e3 − 8. 0 3 After changing the order of integration the integral transforms to e3 1 3 e3 2 dx dy = ln y 1 2x 3 ln y e3 dy = 1 (6 − 2 ln y ) dy The integral ln y dy is computed using integration by parts ln y dy = y ln y − y d(ln y ) = y ln y − y 1 dy = y ln y − y + C y Therefore the integral is e3 1 (6 − 2 ln y ) dy = (6y − 2y ln y + 2y ) π /2 cos x e3 1 = 6e3 − 2e3 (3) + 2e3 − 6 − 2 = 2e3 − 8 8. Do the same things for 0 0 sin x dy dx . Solution. The iterated integral is equal to a double integral over the region bounded by the horizontal line y = 0 , the vertical line x = 0 and the curve y = cos x . The iterated integral is equal to π /2 0 0 π /2 cos x π /2 sin x dy dx = 0 1 (y sin x) u2 2 cos x 0 1 0 dx = 1 , 2 sin x cos x dx = 0 0 u du = = Math 2220 Problem Set 6 Solutions Spring 2010 where in the last integral we used the substitution u = sin x . After changing the order of integration the integral transforms to 1 0 0 1 0 1 cos−1 y 1 sin x dx dy = 0 (− cos x) 1 0 cos−1 y 0 dy = y2 (1 − y ) dy = y − 2 =1− 1 1 =. 2 2 10. For the integral −2 −x x2 −2 (x − y ) dy dx reverse the order of integration to obtain a sum of integrals, and evaluate the resulting integrals. Solution. The integral is over the region D bounded by the parabola y = x2 − 2 and the it into two parts using the horizontal line y = −1 , thus D is union of D1 and D2 where D1 : −√≤ y ≤ −1 √ 2 − y+2≤x ≤ y+2 √ line y = −x . Although this is a type II domain, in order to describe it we need to break −1 ≤ y ≤ 2 √ − y + 2 ≤ x ≤ −y D2 : After changing the order of integration the integral becomes a sum of two integrals −1 −2 y +2 2 √ − y +2 (x − y ) dx dy + −y √ − y +2 −1 (x − y ) dx dy The first integral is equal to −1 −2 √ y +2 √ − y +2 (x − y ) dx dy = √ −1 −2 x2 − xy 2 √ y +2 √ − y +2 dy = −1 −2 −2y y + 2 dy With the substitution u = 1 0 y + 2, y = u2 − 2 , dy = 2u du , this becomes 1 0 −2(u2 − 2)u(2u) du = (−4u4 + 8u2 ) du = 8 u3 4 u5 − 3 5 1 0 = 84 28 −= 35 15 Similarly the second integral is 2 −1 2 −1 −y √ − y +2 2 (x − y ) dx dy = −1 x2 − xy 2 −y √ − y +2 2 −1 dy = 2 y2 y+2 + y2 − −y 2 2 2 1 y2 y3 − −y y + 2 dy = 2 4 3 − 4 4 u3 2 u5 − 5 3 2 1 − y −1 y + 2 dy = 3 − 4 (u2 − 2)u(2u) du = = 3 − 4 139 62 28 =− − 5 3 60 Math 2220 Therefore D Problem Set 6 Solutions (x − y ) dy = 112 − 139 9 28 139 − = =− . 15 60 60 20 1 x 2 Spring 2010 12. Reverse the order of integration in a single integral, and evaluate this integral. 0 0 sin x dy dx + 1 0 2−x sin x dy dx to obtain Solution. The first integral is over the triangle with vertices (0, 0) , (1, 0) and (1, 1) and the second one is over the triangle with vertices (1, 0) , (1, 1) and (2, 0) . Therefore there sum is an integral over the triangles with vertices (0, 0) , (1, 1) and (2, 0) and can be written as 0 1 y 2−y 1 sin x dx dy = 0 (− cos x) 2−y y 1 0 1 dy = 0 cos(y ) − cos(2 − y ) dy = (sin y + sin(2 − y )) = 2 sin 1 − sin 2. sin x dx dy x 0y Solution. The integral is over the triangle with vertices (0, 0) , (π, 0) and (π, π ) . After 16. Evaluate changing the order of integration the integral becomes π 0 0 2 x π 0 π π π sin x dy dx = x 1 y/2 sin x y x x 0 dx = 0 sin x dx = − cos x π 0 =2 18. Evaluate 0 e−x dx dy 2 Solution. The integral is over the triangle with vertices (0, 0) , (1, 0) and (1, 2) . After changing the order of integration, the integral becomes 1 0 1 0 0 2 2x e −x2 1 dy dx = 0 1 0 (ye−x ) 2 2x 0 1 0 dx = 2xe−x dx = 2 e−u du = −e−u = 1 − e−1 , using the substitution u = x Section 5.4 : 2. Evaluate (x2 + y 2 + z 2 ) dV [0,1]×[0,2]×[0,3] 3 0 Solution. The triple integral is equal to the iterated integral 1 0 0 2 (x2 + y 2 + z 2 ) dz dy dx = 0 1 0 2 (x2 z + y 2 z + z 3 /3) 3 0 dy dx = Math 2220 1 0 0 1 0 2 Problem Set 6 Solutions (3x2 + 3y 2 + 9) dy dx = 0 1 Spring 2010 2 0 (3x2 y + y 3 + 9y ) 1 0 dx = (6x2 + 26) dx = 2x3 + 26x = 28. 12. Integrate the function f (x, y, z ) = y over the region bounded by the plane x + y + z = 2 , the cylinder x2 + z 2 = 1 , and y = 0 . Solution. The region of integration is the portion of the interior of the cylinder lying between the xz -plane and the plane x + y + z = 2 , which cuts across the cylinder at an angle. The shadow in the xz -plane of the region of integration is the same as the shadow of the cylinder itself, namely the disk x2 + z 2 ≤ 1 . The domain of integration can be written as −1 ≤ x ≤ 1, and the integral is 1 −1 √ 1−x2 0 √ 1−x2 2−x−z 1 − 1 − x2 ≤ z ≤ 1 − x2 , 0 ≤ y ≤ 2 − x − z, √ − 1−x2 y dy dz dx = −1 √ 1−x2 √ − 1−x2 y2 2 2−x−z 0 dz dx = 1 2 1 2 1 2 1 1 1 −1 (4 √ − 1−x2 2 + x2 + z 2 + 2xz − 4x − 4z ) dz dx = 3 2 2 1−x2 √ − 1−x2 √ −1 (4z + x z + z /3 + xz − 4xz − 2z ) dx = 8 −1 1 − x2 + 2 x2 1 3 1 −1 2 1 − x2 + (1 − x2 )3/2 − 8x 3 1 − x2 (2x2 − 12x + 13) dx 1 − x2 d x = Now substitute x = sin t , so 1 3 1 3 √ 1 − x2 = cos t and dx = cos t dt . This gives π /2 −π/2 π /2 cos t(2 sin2 t − 12 sin t + 13) cos t dt = −π/2 2 sin2 t cos2 t − 12 sin t cos2 t + 13 cos2 t dt Math 2220 Problem Set 6 Solutions Spring 2010 Using the formulas sin2 t = (1 − cos 2t)/2 and cos2 t = (1 + cos 2t)/2 this can be rewritten 1 3 π /2 −π/2 1 − cos 4t 13 − 12 sin t cos2 t + (1 + cos 2t) dt = 4 2 π /2 −π/2 1t sin 4t 13t 13 sin 2t − − 4 cos3 t + − 34 16 2 4 = 9π 1 1 13 π= + 34 2 4 14. Integrate f (x, y, z ) = z over the region in the first octant bounded by the cylinder y 2 + z 2 = 9 and the planes y = x , x = 0 , and z = 0 . Solution. We can look at the shadow in the yz -plane, which is one quarter of the disk y 2 + z 2 ≤ 9 . The domain of integration can then be expressed as 0 ≤ z ≤ 3, which gives the integral 3 0 0 3 0 0 3 0 √ 9−z 2 0 √ 9−z 2 y 0≤y≤ 9 − z2 , 3 √ 9−z 2 0 0≤x≤y z dx dy dz = 0 3 (zx) y 0 dy dz = yz dy dz = 0 y2z 2 √ 9−z 2 0 3 0 dz = 81 8 1 1 9z 2 z4 z (9 − z 2 ) dz = − 2 22 4 = 16. Integrate f (x, y, z ) = 3x over the region in the first octant bounded by z = x2 + y 2 , x = 0 , y = 0 , and z = 4 . Solution. If we look at the shadow in the xy -plane we get the integral 2 0 0 √ 4−x2 4 3x dz dy dx x2 +y 2 This can be computed, but let’s instead do the problem by looking at the shadow in the yz -plane. Then the domain of integration can be expressed as 0 ≤ z ≤ 4, 0≤y≤ √ z, 0≤x≤ √ 0 z − y2 √ 0 which gives the following iterated integral √2 √ 4 z z −y 4 0 z 3x dx dy dz = (3x /2) 2 z −y 2 dy dz = 0 0 0 Math 2220 4 0 0 √ z Problem Set 6 Solutions 3 (z − y 2 ) dy dz = 2 4 0 4 0 Spring 2010 √ z 0 3 (zy − y 3 /3) 2 4 0 dz = z 3/2 dz = 2 5 /2 z 5 = 64 . 5 20. Find the volume of the region inside both the cylinders x2 + y 2 = a2 and x2 + z 2 = a2 . Solution. The part of the region in the first octant is shown in the first figure below. The curved top is part of the cylinder x2 + z 2 = a2 and the curved right side is part of the cylinder x2 + y 2 = a2 . The two cylinders intersect each other when x2 + y 2 = x2 + z 2 , so two symmetric halves, one of which is shown in the second figure. y 2 = z 2 and z = ±y . The plane y = z cuts the part of the region in the first octant into We need to compute the volume of this piece and multiply by 2 to get the volume in the first octant, then multiply by 8 to get the volume in all 8 octants. We can compute the volume of the one piece by looking at its shadow in the yz -plane, which is a triangle. Then the total volume of the region is √2 2 a y 16 0 0 0 a −y a 0 0 y a dx dz dy = 16 a2 − y 2 dz dy = 16 y 0 a2 − y 2 dy 3 /2 a 0 16 2 =− a − y2 3 1 1 0 0 x2 = 16 3 a 3 22. Change the order of integration of equivalent iterated integrals. 0 f (x, y, z ) dz dx dy to give five other Solution. The domain of integration is given by 0≤y≤1 0≤x≤1 0 ≤ z ≤ x2 From the figure we can then read off 5 other iterated integrals equivalent to the original one: Math 2220 Problem Set 6 Solutions Spring 2010 1 0 1 0 1 0 0 0 0 1 0 1 x2 1 x2 0 1 √ z 0 0 1 1 f (x, y, z ) dz dy dx 1 √ 1 0 1 f (x, y, z ) dy dz dx f (x, y, z ) dy dx dz f (x, y, z ) dx dz dy 0 z 1 √ f (x, y, z ) dx dy dz z 2 1 2 √ 24. Consider the iterated integral 0 0 36−9x2 36−4x2 −4y 2 5x2 2 dz dy dx . (a) Describe the region of integration in R3 . Solution. From the inner limits of integration (for dz ) we can deduce that the region lies below the paraboloid z = 5x2 . From the outer two integrals we can deterz = 36 − 4x2 − 4y 2 and above the parabolic cylinder mine that the shadow in the xy -plane is the region in √ the first quadrant below the curve y = 1 36 − 9x2 , 2 which is the ellipse 9x2 + 4y 2 = 36 . The surfaces z = 36 − 4x2 − 4y 2 and z = 5x2 intersect where 36 − 4x2 − 4y 2 = 5x2 , or 9x2 + 4y 2 = 36 , the same ellipse as in the shadow. So the region in R3 is exactly the points in the first octant that lie below the paraboloid 2 dz dx dy . Do not evaluate your answer. z = 36 − 4x2 − 4y 2 and above the parabolic cylinder z = 5x2 . (b) Set up an equivalent triple integral Solution. The shadow in the xy plane is the quarter ellipse 9x2 + 4y 2 ≤ 36 and x, y ≥ 0 . So if all we do is switch dy dx to dx dy we get 3 0 0 1 3 √ 36−4y 2 36−4x2 −4y 2 5x2 2 dz dx dy (c) Set up an equivalent triple integral 2 dy dz dx . Do not evaluate your answer. so the integral is Solution. The shadow in the xz -plane lies between the parabolas z = 5x2 and z = 36−4x2 , 2 0 36−4x2 5x2 0 1 2 √ 36−4x2 −z 2 dy dz dx. ...
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This note was uploaded on 04/21/2010 for the course MATH 2220 taught by Professor Parkinson during the Spring '08 term at Cornell.

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