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Unformatted text preview: Math 2220 Section 5.1 :
9 Problem Set 5 Solutions Spring 2010 e 1 6. Evaluate the integral
1 √ ln x dx dy . xy
e 1 1 0 Solution. The inner integral is
e 1 √ ln x 1 dx = xy 2y ln x 1 dx = x 2y 1 t2 t dt = 2y 2 1 =
0 1 . 4y Thus, the iterated integral is equal to √ 9 e ln x dx dy = xy 1 1 9 1 1 1 dy = ln y 4y 4 9 =
1 ln 9 ln 3 = . 4 2 8. Find the volume of the region bounded on the top by the plane z = x + 3y + 1 , on the bottom by the xy plane, and on the sides by the planes x = 0 , x = 3 , y = 1 , y = 2 . Solution. The volume can be computed as the iterated integral
3 0 1 3 2 3 x + 3y + 1 dy dx =
0 3y 2 xy + +y 2
3 2 dx =
1 =
0 11 x+ dx = 2 11 x2 +x 2 2 =
0 9 33 + = 21. 2 2 For the following two problems, calculate the integral and indicate what region in R3 the integral is computing the volume of.
2 3 10.
0 1 2 dx dy . Solution. The integral represents the volume of the rectangular box (actually a cube) bounded by 1 ≤ x ≤ 3 , 0 ≤ y ≤ 2 and 0 ≤ z ≤ 2 . One can directly see that the volume is equal to 8 , or compute it as the iterated integral
2 0 3 1 0 1 3 2 3 1 2 2 0 2 dx dy =
0 (2x) dy =
0 4 dy = (4y ) = 8. 14.
−2 x sin πy dy dx . Solution. The integral represents the volume of the region bounded by x = −2 , x = 3 , y = 0 , y = 1 , z = 0 , and z = x sin πy . The value of the iterated integral is
3 −2 0 1 3 x sin πy dy dx = −2 −  x cos πy π 1 3 dx =
0 −2 2 x dx. π Math 2220 Problem Set 5 Solutions Spring 2010 This integral involves x and can be computed by breaking the integral into two parts 2 − x dx + π −2 Section 5.2 : In the following two problems evaluate the integral and sketch the region in the xy plane determined by the limits of integration.
2 x2 0 3 0 2 x dx = π x2 − π 0 +
−3 x2 π 2 0 = 9 4 13 += . ππ π 4.
0 0 y dy dx . Solution. The integral is over the region in the ﬁrst quadrant of xy plane, below the parabola y = x2 and to the left of the vertical line x = 2 .
2 0 0 1 x2 2 x2 0 2 y dy dx =
0 y /2 2 dx =
0 x4 x5 dx = 2 10 2 =
0 16 5 √
0 8.
−1 1−y 2 3 dx dy . Solution. The integral is over the half disk inside the circle x2 + y 2 = 1 and to the right of the y axis. We have √2 √ 1 1−y 1 3 dx dy = (3x)
−1 0 −1 0 π 0 1 1−y 2 dy =
−1 0 3 1 − y 2 dy. After the substation y = cos t , this integral becomes −3 sin t dt = −3
2 sin 2t t − 2 4 =
π 3π . 2 12. Integrate the function f (x, y ) = 3xy over the region bounded by y = 32x3 and √ y = x. Solution. The two curves intersect at the points (0, 0) and (1/4, 1/2) . Thus, the region between them is of type 1 and is given by 0 ≤ x ≤ 1 4 √ 3 and 32x ≤ y ≤ x . The double integral is equal to the iterated
1 /4 0 √ x 1 /4 integral 3xy dy dx =
32x3 0 32 xy 2 √ x 1 /4 dx =
32x3 0 32 x − 1536x7 dx 2
1 /4 = x3 − 192x8 2 =
0 3 5 1 − = . 128 1024 1024 Math 2220 14. Evaluate y = 3.
D Problem Set 5 Solutions Spring 2010 3y dA where D is the region bounded by xy 2 = 1 , y = x , x = 0 and Solution. The region D can be split into two type 1 regions or into two type 2 regions. We will use type 2 regions, as shown in the ﬁgure. D1 : 0 ≤ y ≤ 1 and 0≤x≤y 0≤x≤ 1 y2 D2 : 1 ≤ y ≤ 3 and The double integral is 1 y 3 1/y 2 3y dA =
D D1 3y dA +
D2 3y dA =
0 0 3y dx dy +
1 0 3y dx dy The ﬁrst iterated integral is
1 0 0 y 1 3y dx dy =
0 (3xy ) y 0 1 dy =
0 3y 2 dy = y 3 1 0 =1 and the second is
3 1 0 1/y 2 3 3y dx dy =
1 (3xy ) 1/y 2 0 3 dy =
1 3 dy = 3 ln y y 3 1 = 3 ln 3. Therefore the double integral is 1 + 3 ln 3 = 1 + ln 27 . 16. Evaluate
D (x2 + y 2 ) dA where D is the region in the ﬁrst quadrant bounded by y = x , y = 3x , and xy = 3 . Solution. The hyperbola xy = 3 intersects the line y = x at the √√ point ( 3, 3) and it intersects y = 3x at (1, 3) . The region is both type 1 and type 2, but either way we have to break the integral up as the sum of two integrals. If we do it as a type 1 region we get
1 0 3x x 1 √ (x2 + y 2 ) dy dx +
1 3 3/x x √ 3 (x2 + y 2 ) dy dx
3/x x =
0 1 1 x2 y + y 3 3 32 3 x dx + 3 3x x √ 3 1 dx +
1 1 x2 y + y 3 3 dx =
0 8 10 4 =6 3x + 9x−3 − x3 dx = + 3 3 3 Math 2220 Problem Set 5 Solutions Spring 2010 [c, d] , then 28. (a) Show that if R = [a, b] × [c, d] , f is continuous on [a, b] , and g is continuous on
b d f (x)g (y ) dA =
R a f (x) dx
c g (y ) dy Solution. By the Fubini’s Theorem we can write the double integral as an iterated integral
b d b d f (x)g (y ) dA =
R a c f (x)g (y ) dy dx =
a f ( x)
c g (y ) dy dx Here the quantity in parentheses is constant, so it can be pulled outside the dx integral to get
d b b d g (y ) dy
c a f (x) dx =
a f (x) dx
c g (y ) dy . (b) What can you say about
D f (x)g (y ) dA if D is not a rectangle? Speciﬁcally, what if D is an elementary region of type 1? Solution. If D is elementary region we can also use the Fubini’s Theorem to express the double integral as an iterated one. If D is of type 1 we have
b δ (x) b δ (x) f (x)g (y ) dA =
R a γ (x) f (x)g (y ) dy dx =
a f ( x)
γ (x) g (y ) dy dx. However, we can not simplify any further since the limits of integration in the inner integral are not constant, so the quantity in parentheses is not a constant. ...
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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 University (Engineering School).
 Spring '08
 PARKINSON
 Math, Multivariable Calculus

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