Higher_Mathematics_Part_3
.pdfChange the order of integration. Firstly draw the domain of integration.
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y2 +3 |
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3. ∫ dx ∫ f (x, y)dy. |
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4. ∫ dy ∫ f (x, y)dx. |
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3− x |
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1− x2 |
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π / 4 |
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cos x |
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5. ∫ dx ∫ |
f (x, y)dy. |
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6. ∫ dx ∫ f (x, y)dy. |
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sin x |
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x2 |
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8−2x |
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7. ∫ dx ∫ f (x, y)dy + ∫ dx ∫ f (x, y)dy. |
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−1 |
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Calculate the double integral |
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8. ∫ dx∫ (x2 + y2 )dy. |
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9. ∫ dx∫ |
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(x + y) |
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Evaluate the double integrals ∫∫ f (x, y)dxdy. |
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10. ∫∫ (x3 + 4y)dxdy, over the domain D bounded by the lines |
x = 0, |
x = 1 , |
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D |
y = 1 . |
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y = 0 and |
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11. ∫∫ |
x |
dxdy, over the domain D bounded by the lines x = 2, |
y = x and |
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D |
y |
xy = 1. |
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the hyperbola |
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12. ∫∫ ydxdy, |
over the domain D bounded by the lines y = 0, |
x + y = 2 |
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D |
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y = |
x . |
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and the parabola |
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Evaluate the following integrals using polar coordinates system. |
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13. ∫∫ |
25 − x2 − y2 dxdy, |
over |
the domain |
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D bounded by |
the |
circle |
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D |
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x2 + y2 = 9 . |
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14. ∫∫ (x2 + y2 )2 dxdy, if the domain D is the annulus 1 ≤ x2 + y2 ≤ 4. |
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D |
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15. ∫∫ |
4 − x2 − y2 dxdy, |
over a domain D bounded by the circle |
x2 + y2 = 4 |
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D |
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and lines y = x , |
y = |
3x ( x > 0, y > 0 ). |
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16. ∫∫ ydxdy, |
over a domain |
D bounded |
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by the |
circles |
x2 + y2 |
= 4x, |
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D |
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x2 + y2 = 8x and lines |
y = |
3x , x = 3y. |
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Calculate the area bounded by the lines.
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x2 |
= 4y + 4, |
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= −2 y + 4. |
18. |
x2 |
+ y2 = 4, |
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= 3x (x ≥ 0). |
Using the geometrical interpretation of the double integral, calculate the volume of a cylindrical body bounded by the following surfaces
19. x2 + y2 + z = 4, z = 0.
20. |
z = x2 + y2 , x = 0, |
x = 1, |
y = 0, |
y = 2, |
z = 0. |
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21. |
z = 0, |
y + z = 2, |
y = x2 . |
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22. |
Calculate the volume of a body bounded by the cylinder |
x2 + y2 = Rx |
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and sphere x2 + y2 + z2 |
= R2 . |
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23. |
Find |
the mass |
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plate D bounded |
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curves |
x2 + y2 = 4, |
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x2 + y2 |
= 16 ( x ≥ 0, y ≥ 0 ), |
if |
the |
density |
in each |
point |
(x, y) D equals |
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γ(x, y) = |
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x2 + y2 |
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Find the centre of mass of the homogeneous figures bounded by the following lines
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24. |
y = x2 + 1, |
x − y + 3 = 0. |
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25. |
x2 + y2 = 4, x = y, |
y = |
3x ( x ≥ 0, |
y ≥ 0 ). |
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Answers |
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4 x |
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4 y |
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x+5 |
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1. ∫dx∫ f (x, y)dy |
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∫dy ∫ f (x, y)dx + ∫dy∫ f (x, y)dx. |
2. ∫ dx ∫ f (x, y)dy + |
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4 y−2 |
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+∫dx ∫ f (x, y)dy |
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∫dy ∫ f (x, y)dx + ∫dy ∫ f (x, y)dx. |
3. ∫dy ∫ f (x, y)dx + |
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1 x+ 2 |
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4 y−5 |
0 3− y |
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+∫dy∫ f (x, y)dx. |
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4. ∫dx ∫ f (x, y)dy + ∫dx ∫ f (x, y)dy + ∫dx ∫ f (x, y)dy. |
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x−3 |
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x−3 |
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1− y2 |
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arcsin x |
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arccos x |
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5. ∫dy |
∫ f (x, y)dx. 6. ∫ |
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∫ f (x, y)dx + ∫ dy ∫ f (x, y)dx. 7. ∫ dy∫ f (x, y)dx + |
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+∫dy |
∫ |
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f (x, |
y)dx. |
8. |
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9. |
0,5. |
10. |
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11. |
. 12. |
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13. |
π. |
14. |
21π. |
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15. 12( |
3 − 1). |
16. |
56 |
. 17. 8. 18. (4π + |
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3) / 3 . 19. 8π. |
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21. |
32 |
2 |
. 22. |
2(3π − 4) |
R3. |
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23. 6. |
24. (0,5; 2,6). 25. |
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− 2); |
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2 − 1) |
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Micromodule 5
SELF-TEST ASSIGNMENTS
5.1. Change the order of integration. Firstly, draw the domain of integration.
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x+1 |
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5.1.1. а) |
∫ dx |
∫ |
f (x, y)dy; |
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−1 |
x2 −1 |
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x+ 2 |
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5.1.2. а) |
∫ dx ∫ |
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f (x, y)dy; |
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2x+1 |
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5.1.3. а) |
∫ dx |
∫ |
f (x, y)dy; |
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x2 +1 |
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5.1.4. а) |
∫ dx |
∫ |
f (x, y)dy; |
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5.1.5. а) |
∫ dx ∫ |
f (x, y)dy; |
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2− x2 |
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5.1.6. а) |
∫ dx |
∫ |
f (x, y)dy; |
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x+ 2 |
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5.1.7. а) |
∫ dx ∫ |
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f (x, y)dy; |
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5.1.8. а) |
∫ dx |
∫ |
f (x, y)dy; |
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4− x2 |
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5.1.9. а) |
∫ dx |
∫ |
f (x, y)dy; |
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b) ∫ dx |
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∫ |
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f (x, y)dy . |
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−2− |
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4x− x2 |
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9−8x− x2 |
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b) ∫ dx |
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∫ |
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f (x, y)dy . |
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b) ∫ dx |
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∫ |
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f (x, y)dy . |
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3+ 2x− x2 |
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16− x2 |
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b) ∫ dx ∫ |
f (x, y)dy . |
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4− 3+ 2x− x2 |
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b) ∫ dx |
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∫ |
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f (x, y)dy . |
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b) ∫ dx ∫ |
f (x, y)dy. |
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b) ∫ dx |
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∫ |
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f (x, y)dy . |
0−2− 4x− x2
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b) |
∫ dx |
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∫ |
f (x, y)dy . |
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0 − |
9−8x− x2 |
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−1+ −6x− x2 |
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b) |
∫ dx |
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∫ |
f (x, y)dy . |
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5.1.10. а) ∫ dx |
∫ f (x, y)dy; |
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x2 −3 |
12− x
5.1.11.а) ∫ dx ∫ f (x, y)dy;
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5.1.12. а) |
∫ dx ∫ |
f (x, y)dy; |
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2 x |
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5.1.13. а) |
∫ dx ∫ |
f (x, y)dy; |
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5.1.14. а) |
∫ dx ∫ |
f (x, y)dy; |
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5.1.15. а) |
∫ dx ∫ f (x, y)dy; |
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x+ 2 |
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5.1.16. а) |
∫ dx ∫ |
f (x, y)dy; |
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5.1.17. а) |
∫ dx∫ f (x, y)dy; |
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5.1.18. а) |
∫ dy ∫ |
f (x, y)dx; |
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5.1.19. а) |
∫ dy ∫ |
f (x, y)dx; |
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5.1.20. а) |
∫ dx ∫ |
f (x, y)dy; |
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2x |
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5.1.21. а) |
∫ dx ∫ |
f (x, y)dy; |
1x
12− x2
5.1.22.а) ∫ dx ∫ f (x, y)dy;
−1 x2
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3+ −2x− x2 |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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6x− x2 −5 |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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10x− x2 |
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6x− x2 |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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36− x2 |
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b) ∫ dx ∫ |
f (x, y)dy . |
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−1 1+ 6x+ x2 −5 |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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2x− x2 |
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3− −2x− x2 |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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9+8x− x2 |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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6x− x2 −5 |
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2x− x2 |
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b) ∫ dx |
∫ |
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f (x, y)dy . |
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1x2
5.1.23.а) ∫ dx ∫ f (x, y)dy;
−1 |
x−1 |
2 |
3x−2 |
5.1.24. а) ∫ dx |
∫ f (x, y)dy; |
1 |
1 |
x
1 |
y2 +1 |
5.1.25. а) ∫ dy |
∫ f (x, y)dx; |
0− y2
1y+1
5.1.26. а) |
∫ dy ∫ |
f (x, y)dx; |
||
|
0 |
− y |
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|
|
2 |
|
2x |
|
5.1.27. а) |
∫ dx |
|
∫ |
f (x, y)dy; |
|
0 |
x2 −1 |
||
|
0 |
x2 + 2 |
||
5.1.28. а) |
∫ dx |
∫ |
f (x, y)dy; |
|
|
−1 |
|
1 |
|
|
4 |
2 |
x |
|
5.1.29. а) |
∫ dx ∫ |
f (x, y)dy; |
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1 |
|
x |
|
43
5.1.30.а) ∫ dx ∫ f (x, y)dy;
1x
0 |
|
|
x+ 4 |
|
b) ∫ dx ∫ |
f (x, y)dy . |
|||
−4 |
− |
16− x2 |
|
|
−3 |
|
|
2 |
|
b) ∫ dx |
∫ |
f (x, y)dy . |
||
−5 |
− |
6x+ x2 −5 |
||
3 |
3+ |
|
3+ 2x− x2 |
|
b) ∫ dx |
|
|
∫ |
f (x, y)dy . |
1 |
|
|
0 |
|
3 |
2+ |
|
6x− x2 −5 |
|
b) ∫ dx |
|
|
∫ |
f (x, y)dy . |
1 |
|
|
0 |
|
−1 |
|
|
3 |
|
b) ∫ dx |
∫ |
f (x, y)dy . |
||
−2 |
2− |
−2x− x2 |
||
0 |
1+ |
−6x− x2 |
||
b) ∫ dx |
∫ |
f (x, y)dy . |
||
−3 |
|
|
0 |
|
5 |
|
|
0 |
|
b) ∫ dx |
|
|
∫ |
f (x, y)dy . |
1 |
− |
10x− x2 |
|
|
0 |
|
|
2 |
|
b) ∫ dx |
∫ |
f (x, y)dy . |
||
−1 |
− |
9+8x− x2 |
5.2. Calculate the double integral ∫∫ f (x, y)dxdy over the domain D.
D
5.2.1. ∫∫ (x + 2 y + 1)dxdy , over the domain D bounded by the lines: y = x2 ,
D
y = 1 , x = 0 , y = 0 .
5.2.2. ∫∫ (x2 + y2 )dxdy , over the domain D bounded by the lines: y = x − 1,
D
y = 3 − 2x , x = 0 .
5.2.3. ∫∫ (9 − y2 )dxdy , over the domain D bounded by the lines: y = 4 − 2x ,
D
y = 2x , y = 0 .
5.2.4. ∫∫ (x − y + 3)dxdy , over the domain D bounded by the lines: y = x ,
D
y = (x − 2)2 , y = 0 .
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5.2.5. ∫∫(2x + 1)dxdy , over the domain D bounded by the lines: x = 4 , y2 = x.
D
5.2.6. ∫∫(x − 4y)dxdy , over the domain D bounded by the lines: y = x − 4 ,
|
D |
|
|
y = x , |
y = 0 . |
|
|
5.2.7. |
∫∫ (x2 − 2y)dxdy , over the domain D bounded by the lines: |
x = 1 , |
|
|
D |
= 0 . |
|
x = 2 , y = x , y |
|
||
5.2.8. |
∫∫ (x2 + 4 y)dxdy, over the domain D bounded by the lines: |
x = 2 , |
|
|
D |
|
|
y = 2x , 2y = x . |
|
|
|
5.2.9. |
∫∫ (2x − 3y2 )dxdy , over the domain D bounded by the lines: |
y = 2 , |
|
|
D |
|
|
y = 2x , 2y = x . |
|
|
5.2.10. ∫∫ (x + 2 y −1)dxdy , over the domain D bounded by the lines: x = 1 ,
D
x = 0 , y = x2 , x + y = 0 .
5.2.11. ∫∫ 2xydxdy , over the domain D bounded by the lines: y = 2 ,
D
y = 1− x , y = x − 1.
5.2.12. ∫∫ (2y − x + 2)dxdy , over the domain D bounded by the lines: y = 1,
D
y= 2 − x2 .
5.2.13.∫∫ ( y − 2x + 6)dxdy , over the domain D bounded by the lines: y = 0 ,
D
y = x , y = 2 − x .
5.2.14. ∫∫(3x2 − y2 )dxdy, over the domain D bounded by the lines: y = x + 1, y = 1, y = xD, y = 0 .
5.2.15. ∫∫ (2x + 3)dxdy , over the domain D bounded by the lines: x = 0 ,
D
y= 1− x2 , y = −1− x .
5.2.16.∫∫ (x + 3y2 )dxdy , over the domain D bounded by the lines: y = x + 1,
x= −1 , y =Dx , x = 1 .
5.2.17.∫∫ (2x + y + 2)dxdy , over the domain D bounded by the lines: x = 0 ,
D
y = 0 , y = x2 + 1 , x = 1 .
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5.2.18. ∫∫ ( x + 2y)dxdy, over the domain D bounded by the lines: y = x +1,
D
x = 0 , y = 0 , x = 1 .
5.2.19. ∫∫ (x − y )dxdy , over the domain D bounded by the lines: x = 2 ,
D
y = 1 , y = 0 , y = − x .
5.2.20. ∫∫ (2xy − 1)dxdy , over the domain D bounded by the lines: y = x + 1,
D
y = 1 , y = 0 , x = 1 .
5.2.21. ∫∫ (4x − y + 1)dxdy , over the domain D bounded by the lines:
D
y = − x2 , x = 0 , y = 1 , x = −1 .
5.2.22. ∫∫ (3 x + 4y)dxdy , over the domain D bounded by the lines: y = −2 ,
D
y = x , y = − x .
5.2.23. ∫∫ (x − 2y + 5)dxdy , over the domain D bounded by the lines: y = 0 ,
D
y = 1 , y = − x , x = 2 .
5.2.24. ∫∫ (x + 1) ydxdy , over the domain D bounded by the lines: y = 0,
D
y = x, y = x − 1, x = 2 .
5.2.25. ∫∫ (3x + y + 2)dxdy , over the domain D bounded by the lines: x = 1 ,
D
y = 2x , y = − x .
5.2.26. ∫∫ (x + 3y2 + 1)dxdy , over the domain D bounded by the lines: y = 3 ,
D
y = x2 −1 . |
|
5.2.27. |
∫∫ (x2 + 2xy + 2)dxdy , over the domain D bounded by the |
lines: y = 0 , |
D |
y = 2 − x , y = 1, x = 0 . |
5.2.28. ∫∫ (4x − 2y + 1)dxdy , over the domain D bounded by the lines: y = 0 ,
D
y = x2 , x = −2 . |
|
|
|
|
5.2.29. ∫∫ (xy + 1)dxdy , over the domain D bounded by the lines: |
y = |
1 |
, |
|
x |
||||
D |
|
|
y = 1, y = 2 , x = −1 .
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5.2.30. ∫∫ (x + 4y + 2)dxdy , over the domain D bounded by the lines:
|
1 |
|
D |
|
y = |
, |
y = 2 , y = x . |
||
x |
||||
|
|
|
5.3. Calculate the area of a domain D bounded by the following lines. Evaluate the following integrals using polar coordinates system.
5.3.1. |
x2 + y2 + 6x ≤ 0, |
y ≥ |
3x, |
y ≤ 0. |
||
5.3.2. |
x2 |
+ y2 |
+ 4 y ≤ 0, |
y ≤ |
3x, |
x ≤ 0. |
5.3.3. |
x2 |
+ y2 |
− 6x ≤ 0, |
y ≤ |
3x, |
y ≥ − x. |
5.3.4. |
x2 |
+ y2 |
− 4y ≤ 0, |
y ≥ |
3x, |
x ≥ 0. |
5.3.5. y ≥ 0, |
y ≤ |
3x, |
x2 + y2 |
≤ 10x . |
|||||||
5.3.6. x2 + y2 |
≤ 8y, |
3y ≥ x, |
y ≤ |
3x. |
|||||||
5.3.7. y ≥ − x, |
|
y ≤ − |
3x, |
x2 + y2 − 4y ≤ 0. |
|||||||
5.3.8. x2 + y2 + 10x ≤ 0, |
3y ≤ x, |
y ≥ x. |
|||||||||
5.3.9. y ≥ 0, |
y ≤ x, |
x2 + y2 − 6x ≤ 0. |
|||||||||
5.3.10. x2 |
+ y2 |
≤ 2y, |
y ≤ x, |
3y ≥ x. |
|||||||
5.3.11. |
3y ≥ x, |
y ≤ − x, |
x2 + y2 + 4x ≤ 0. |
||||||||
5.3.12. |
x2 |
+ y2 |
+ 10x ≤ 0, |
y ≥ x, |
y ≤ 0. |
||||||
5.3.13. |
y ≥ 0, |
|
y ≤ |
3x, |
x2 + y2 |
≤ 4x. |
|||||
5.3.14. |
x2 |
+ y2 |
≤ 2y, |
y ≤ |
3x, |
|
y ≤ − x. |
||||
5.3.15. |
x2 |
+ y2 |
+ 4x ≤ 0, |
3y ≥ − x, |
y ≤ − 3x. |
||||||
5.3.16. |
x2 |
+ y2 |
+ 16y ≤ 0, |
y ≤ |
3x. |
|
|||||
5.3.17. |
y ≤ x, |
|
3y ≥ x, |
y2 ≤ x(6 − x). |
|||||||
5.3.18. |
x ≥ 0, |
|
y ≥ |
3x, |
x2 + y2 |
≤ 8y. |
|||||
5.3.19. |
x2 |
+ y2 + 4x ≤ 0, |
3y ≤ − x, |
y ≥ 0. |
|||||||
5.3.20. |
x2 |
+ y2 + 10 y ≤ 0, |
y ≥ x. |
|
|
||||||
5.3.21. |
3y ≥ − x, |
y ≤ 0, |
y2 ≤ x(8 − x). |
5.3.22.3y ≤ − x, x2 + y2 ≤ 8y.
5.3.23. |
x2 |
+ y2 |
+ 12 y ≤ 0, |
y ≥ 3x, |
3y ≤ x. |
5.3.24. |
x2 |
+ y2 |
+ 2 y ≤ 0, |
y ≤ − x, y ≤ |
3x. |
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5.3.25. |
y ≥ 3x, x2 + y2 |
≤ 6x. |
|
|||
5.3.26. |
y ≤ − x, |
3y ≥ − x, |
x2 + y2 ≤ 8y. |
|||
5.3.27. |
x2 |
+ y2 + 4x ≥ 0, |
y ≤ 0, |
3y ≥ x. |
||
5.3.28. |
y ≥ − x, |
y ≤ − 3x, |
x2 + y2 ≤ 8y. |
|||
5.3.29. |
x2 |
+ y2 |
+ 4x ≤ 0, |
|
3y ≤ x, |
y ≥ x. |
5.3.30. |
x2 |
+ y2 |
+ 16 y ≤ 0, |
|
3y ≤ − x, y ≤ 3x. |
5.4. Using the geometrical sense of the double integral, calculate the volume of a cylindrical body bounded by the following surfaces.
5.4.1. x = |
y, |
|
y = 4, |
x = 0, |
z = 0, |
x + z = 6. |
||||||
5.4.2. z = 4 − x2 , |
x + y − 2 = 0, x = 0, y = 0, |
z = 0. |
||||||||||
5.4.3. y = x2 , |
|
y = 0, |
x = 2, |
z = 4 − x2 , |
z = 0. |
|||||||
5.4.4. x + y = 2, |
x = 0, z = x2 + y2 , |
x = |
y, |
z = 0. |
||||||||
5.4.5. z = 1+ x2 , |
x + y − 2 = 0, x = 0, |
y = 0, z = 0. |
||||||||||
5.4.6. y = x2 , |
y = 0, |
x + y = 2, |
z = 0, z + y = 3. |
|||||||||
5.4.7. z = x2 , |
|
x + y − 2 = 0, |
y = 0, z = 0. |
|
||||||||
5.4.8. y = x, |
y = 2x, |
y = 2, |
z = 6 − y2 , |
z = 0. |
||||||||
5.4.9. y = |
x, |
|
x = 4, |
x + 2 y = 0, z − x = 2, z = 0. |
||||||||
5.4.10. z = 1+ x2 , |
x = 2, |
x + y − 4 = 0, y = 0, |
z = 0. |
|||||||||
5.4.11. x = y, |
2y + x = 0, |
x = 2, z = x2 + y2 , |
z = 0. |
|||||||||
5.4.12. y = 2x, |
|
y + 2x = 4, |
y = 0, |
z = x2 , |
z = 0. |
|||||||
5.4.13. y = x2 , |
|
y = 1, |
z = y2 , |
z = 0. |
|
|
||||||
5.4.14. y = 2x, |
x = 2, |
y = 1, |
z = y2 , |
z = 0. |
|
|||||||
5.4.15. x − 2y = 0, |
z = 4 − x2 , y = 0, |
z = 0. |
|
|||||||||
5.4.16. z = y2 , |
|
x + y = 6, |
y = x, z = 0. |
|
||||||||
5.4.17. z = 4 − x2 , |
x + y − 4 = 0, |
y = x, |
x = 0, |
z = 0. |
||||||||
5.4.18. z = 4 + y2 , |
x + y − 4 = 0, |
x = 0, |
y = 0, |
z = 0. |
||||||||
5.4.19. z = 4 − y2 , |
x + y − 4 = 0, |
y = x, |
y = 0, z = 0. |
|||||||||
5.4.20. z = y2 , |
x + y − 6 = 0, y = x, z = 0. |
|
||||||||||
5.4.21. y = |
x, |
|
y + x = 0, x = 1, z = x2 + y2 , |
z = 0. |
||||||||
5.4.22. y = |
x, |
|
x = 0, |
y = 2, |
z = x2 + 1, z = 0. |
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5.4.23. |
y = x2 , |
y = 4, x = 1, z = 2 + x, |
z = 0. |
||||
5.4.24. |
y = x + 1, |
x = 1, |
y = 0, |
z = 1+ y2 , z = 0. |
|||
5.4.25. |
y = x, y = x + 1, |
x = 0, |
x = 2, |
z = x2 + y2 , z = 0. |
|||
5.4.26. |
z = 2 + x2 , |
x + y − 1 = 0, |
x = 0, |
x − y = 1, z = 0. |
|||
5.4.27. |
y = x2 , y + x = 0, x = 1, |
z = 0, |
z + x = 3. |
||||
5.4.28. |
y = 4 − x2 , |
y = 0, |
z = y, z = 0. |
||||
5.4.29. |
x = 1− y2 , |
z = |
x, |
z = 0. |
|
||
5.4.30. |
x = 2, z = |
y, |
y = x, z = 0. |
|
Micromodule 6
BASIC THEORETICAL INFORMATION.
TRIPLE INTEGRALS
Triple integrals. Principal concepts and definitions. Conditions of the existence and properties. Cylindrical and Spherical coordinates. Change of variable in double integral. Application of double integrals.
Key words: triple integral — подвійний інтеграл; integral sum — інтегральна сума; iterated (repeated) integral — повторний інтеграл; rectangular parallelepiped — прямокутний паралелепіпед; nonhomogeneous material plate —
неоднорідна матеріальна пластина, mass of a planar figure — маса плоскої фігури; moment of inertia — момент інерції.
Literature: [3, chapter 2, section 2.3], [9, chapter 10, § 2], [15, chapter 12, section 12.2], [16, chapter 14, § 11—14], [17, chapter 2, § 8].
6.1. Principal Definitions. Condition for the Existence of a Triple Integral
A triple integral is an analogue of a double integral and is introduced for a function of three variables.
Suppose we are given a material body occupying a three-dimensional region
Ωfilled with a matter. The density at each point of it is know too. Suppose
μ= μ(P) = μ(x, y, z) .
The task is required to find the mass m of the body.
We will break Ω into nonoverlaping cubable (i.e., having a volume) parts Ωi , i = 1, 2, 3… n having volumes Vi , respectively.
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