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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 Spring '08
 PARKINSON
 Integrals, Multivariable Calculus, dx, dy

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