Iterated Integrals and Area (Exercises)
- Last updated
- Jun 14, 2019
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- Page ID
- 21139
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Terms and Concepts
1. When integrating fx(x,y) with respect to x, the constant of integration C is really which: C(x) or C(y)? What does this mean?
- Answer:
- The constant of integration will be C(y) since y is considered a constant in the integration and thus, the original function f could have terms that are functions of y that were lost when taking the partial derivative with respect to the variable x.
2. Integrating an iterated integral is called _________ __________.
- Answer:
- iterated integration or multiple integration
3. When evaluating an iterated integral of the form ∫ba∫g2(x)g1(x)dydx, we integrate from _______ to ________, then from _________ to __________.
- Answer:
- We integrate from bottom to top, then from left to right. Here, from y=g1(x) to y=g2(x), then from x=a to x=b.
More generally, we integrate from curve to curve and then from point to point.
4. One understanding of an iterated integral is that ∫ba∫g2(x)g1(x)dydx gives the _______ of a plane region.
- Answer:
- area
Problems
In Exercises 5-10, evaluate the integral and subsequent iterated integral.
5. (a) ∫52(6x2+4xy−3y2)dy
(b) ∫2−3∫52(6x2+4xy−3y2)dydx
- Answer:
- (a) ∫52(6x2+4xy−3y2)dy =(6x2y+2xy2−y3)|52 =30x2+50x−125−(12x2+8x−8) =18x2+42x−117
(b) ∫2−3∫52(6x2+4xy−3y2)dydx =∫2−3(18x2+42x−117)dx =(6x3+21x2−117x)|2−3 =48+84−234−(−162+189+351) =−102−378 =−480
6. (a) ∫π0(2xcosy+sinx)dx
(b) ∫π/20∫π0(2xcosy+sinx)dxdy
7. (a) ∫x1(x2y−y+2)dy
(b) ∫20∫x1(x2y−y+2)dydx
- Answer:
- (a) ∫x1(x2y−y+2)dy =(x2y22−y22+2y)|x1 =x42−x22+2x−(x22−12+2) =x42−x2+2x−32
(b) ∫20∫x1(x2y−y+2)dydx =∫20(x42−x2+2x−32)dx =(x510−x33+x2−32x)|20 =3210−83+4−3−0 =9630−8030+3030 =2315
8. (a) ∫y2y(x−y)dx
(b) ∫1−1∫y2y(x−y)dxdy
9. (a) ∫y0(cosxsiny)dx
(b) ∫π0∫y0(cosxsiny)dxdy
- Answer:
- (a) ∫y0(cosxsiny)dx =(sinxsiny)|y0 =sin2y
(b) ∫π0∫y0(cosxsiny)dxdy =∫π0sin2ydy =∫π01−cos2y2dy =y−12sin2y2|π0 =π2−0 =π2
10. (a) ∫x0(11+x2)dy
(b) ∫21∫x0(11+x2)dydx
- Answer:
- (a) ∫x0(11+x2)dy =y1+x2|x0 =x1+x2
(b) ∫21∫x0(11+x2)dydx =∫21(x1+x2)dx =12ln(1+x2)|21 =12ln(5)−12ln(2) =12ln(52)
In Exercises 11-16, a graph of a planar region R is given. Give the iterated integrals, with both orders of integration dydx and dxdy, that give the area of R. Evaluate one of the iterated integrals to find the area.
11.
- Answer:
- Area=∫41∫1−21dydx =∫1−2∫411dxdy
∫1−2∫411dxdy =∫1−2x|41dy =∫1−23dy =3y|1−2 =3(3) =9units2
12.
13.
- Answer:
- Area=∫42∫7−xx−11dydx =∫31∫y+121dxdy+∫53∫7−y21dxdy
∫42∫7−xx−11dydx =∫42y|7−xx−1dx =∫42(8−2x)dx =(8x−x2)|42=32−16−16+4 =4units2
14.
15.
- Answer:
- Area=∫10∫√xx41dydx =∫10∫4√yy21dxdy
∫10∫√xx4dydx =∫10y|√xx4dx =∫10(√x−x4)dx =(23x3/2−x55)|10 =23−15 =1015−315 =715units2
16.
In Exercises 17-22, iterated integrals are given that compute the area of a region R in the xy-plane. Sketch the region R, and give the iterated integral(s) that give the area of R with the opposite order of integration.
17. ∫2−2∫4−x20dydx
18. ∫10∫5−5x25−5xdydx
19. ∫2−2∫2√4−y20dxdy
20. ∫3−3∫√9−x2−√9−x2dydx
21. ∫10∫√y−√ydxdy+∫41∫√yy−2dxdy
22. ∫1−1∫(1−x)/2(x−1)/2dydx