Higher Mathematics. Part 3
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y = |
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– 1 |
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– 1 |
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x = y2 |
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Fig. 5.10 |
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Therefore, we can apply here (5.20) |
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∫∫(x + 2y)dxdy = ∫ dx ∫ (x + 2 y)dy = ∫(xy + y2 ) |
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x2x dx = |
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D |
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0 x2 |
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= ∫(x x + x − x3 − x4 )dx = ( |
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I = ∫∫(x + 2y)dxdy |
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Example 2. We take the double integral |
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over a domain |
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y = 2 − x and y = x2 . |
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D |
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D bounded by the lines |
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Solution. Our domain is regular in both directions (Fig. 5.11). Let’s find coordinates of intersection points of parabola y = x2 and straight lines y = 2 − x.
We have x2 = 2 − x, |
then x2 + x − 2 = 0. We obtain x |
= −2, x = 1. |
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y = x2 |
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y = 2 – x |
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Fig. 5.11 |
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111
We obtain
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2− x |
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2− x |
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I = ∫∫ (x + 2y)dxdy = ∫ dx ∫ |
(x + 2y)dy = ∫ dx(xy + y2 ) |
= |
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D |
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−2 |
x2 |
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−2 |
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(4 − 2x − x3 − x4 )dx = |
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= ∫ |
(x(2 − x) + (2 − x)2 − x3 − x4 )dx = ∫ |
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−2 |
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−2 |
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4x |
− x2 |
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32 |
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= 4 − 1− |
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−8 − 4 − 4 + |
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= 12,15. |
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4 5 |
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Example 3. Let’s change the order of integration. We preliminarily draw the domain of integration: ∫∫ f (x, y)dxdy over the domain D bounded by the lines
D
y = 2x, x + y = 3, y = 0 .
Solution. Examine D. The straight lines y = 2x and y = 3 − x meet at point A(1, 2). Domain D is a triangle (Fig. 5.12).
1. This domain is regular. So, y varies from 0 to 2, and |
ϕ1 (x) = 2x and |
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ϕ2 (x) = 3 − x. |
Any straight line |
y = const, (0 ≤ y ≤ 2) |
meets the boundary of |
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the region at not more than two points. |
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Therefore, we can apply here |
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∫∫ f (x, |
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3− x |
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y)dxdy = ∫ dy ∫ f (x, |
y)dy. |
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D |
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2x |
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y = 2x |
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А |
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y = 3 – x |
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О |
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Fig. 5.12 |
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2. Let’s change the order of the integration. The domain should be broken
into two parts, each of which will be regular: if |
0 ≤ x ≤ 1, then 0 ≤ y ≤ 2x ; if |
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1 ≤ x ≤ 3 , then 0 ≤ y ≤ 3 − x . We obtain |
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∫∫ f (x, |
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2 x |
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3− x |
y)dxdy = ∫ dx ∫ f (x, |
y)dy + ∫ dx ∫ f (x, y)dy. |
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D |
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112 |
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Example 4. Take the double integral ∫∫ f (x, y)dxdy, if
D
f (x, y) = x + 2y over a domain D is bounded by the lines
x = 0, y = |
x |
and y = 5 − x2 . |
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2 |
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Solution. We have the domain D bounded by straight |
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lines x = 0, |
y = |
x |
and the parabola |
y = 5 − x2 (Fig. 5.13). |
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x varies from 0 to 2, and ϕ (x) = |
and ϕ |
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(x) = 5 − x2 . |
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y
5
y = 5 – x2
D |
х = 2у |
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1
О 2 x
Fig. 5.13
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5− x2 |
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Therefore ∫∫ f (x, |
y)dxdy = ∫ dx ∫ |
(x + 2 y)dy = ∫(xy + 2 |
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43 |
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= ∫(x(5 − x2 ) + (5 |
− x2 )2 − |
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)dx = ∫(x4 − x3 − |
x2 |
+ |
5x + 25)dx = |
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+ 25x |
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− 4 − |
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+ 10 |
+ 50 ≈ 34. |
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For the replacement of integration order we shall construct area D on Оy, and receive a segment [0,5] . As the regular boundary of the domain consists of two
different lines, then we shall divide this domain into two parts by line |
y = 1. |
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Defined from the equations |
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y = |
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and |
y = 5 − x2 |
a variable y through |
x, we |
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shall receive |
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2 y |
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5− y |
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∫∫ f (x, y)dxdy = ∫ dy ∫ (x + 2 y)dx + ∫ dy ∫ (x + 2 y)dx = |
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D |
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1 x2 |
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2 y |
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5 x2 |
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5− y |
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= ∫ |
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+ 2yx |
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dy + ∫ |
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+ 2 yx |
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dy ≈ −21.56. |
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Example 5. Calculate the integral |
∫∫(6x − 3y)dxdy, where D :{ x + y = 1, |
D
x + y = 2, 2x − y = 1, 2x − y = 3} .
Solution. Let’s construct the given domain D which is bounded by straight lines y = 1− x, y = 2 − x, y = 1+ 2x, y = 3 + 2x (Fig. 5.14).
113
We substitute
x + y = u, 2x − y = v.
Therefore all equations of the straight lines can be written as
x + y = 1 into u = 1, x + y = 2 into u = 2, 2x − y = 1 into v = 1, 2x − y = 3 into v = 3.
Let’s construct the domain |
D* : |
{ |
} |
in uv-plane (Fig. 5.15). |
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u = 1, u = 2, v = 1, v = 3 |
Further for a determination of a Jacobian, it is necessary to transform it so that to receive x and y through u and v. We have
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x = |
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3x = u + v |
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x + y = u |
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2x − y = v |
2x − y = v |
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y = |
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(2u − v) |
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∂x |
Now we find partial derivatives of the first order with respect to u and v, |
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∂x = |
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∂y |
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∂y |
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1 |
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. Then Jacobian equals |
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∂v |
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∂u 3 |
∂v 3 |
∂u 3 |
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3 |
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Fig. 5.15 |
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We obtain |
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Now the given integral is equal to |
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∫∫(6x − 3y)dxdy = ∫∫[6 |
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(2u − v)] |
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dudv = |
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∫∫3vdudv =∫ du∫vdv = ∫ |
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Example 6. Take the |
∫∫ |
1 |
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dxdy |
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where D : |
{ |
x2 + y2 ≤ 1, x ≥ 0, y ≥ 0 |
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is the quarter of the unit circle in the first quadrant.
Solution. Let’s consider a domain D (Fig. 5.16). We pass to polar coordinates
x = ρ cos ϕ and y = ρ sin ϕ, |
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* |
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≤ ρ ≤ 1; 0 |
≤ ϕ ≤ |
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and our integrable |
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function f (x, y) = |
1 |
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= |
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1 |
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= |
1 |
, |
then the given |
integral can be |
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Example 7. Find the area S if D: x = y2 − 2 y, x − y = 0. |
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Solution. We write the given equation of |
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y2 − 2y + 1−1 = x, then we have ( y − 1)2 = x + 1 . It is parabola with centre in the point A(–1; 1). Second equation x − y = 0 is a straight line (Fig. 5.17).
115
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S = ∫∫ dxdy = ∫ dy |
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Example 8. Find the area S if D: x2 + y2 |
= 4x, |
y = x, y = 0. |
Solution. We construct the given figure (Fig. 5.18). The equation x2 + y2 = 4x is a circle of radius 2 with the centre at the point A(2; 0). Let’s find perfect square with respect to x, we obtain
x2 − 4x + y2 = 0, (x2 − 4x + 4) + y2 = 4, (x − 2)2 + y2 = 4.
The graph of the equations y = x and y = 0 are straight lines. They together
with the arc of a circle determine the domain D, which is a curvilinear sector with a top in the origin of coordinates.
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A(2; 0) |
Fig. 5.18
Taking into account the form of the domain, we shall calculate a double integral in a polar system.
116
In polar coordinates the equation of the circle can be written as
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ρ2 cos2 ϕ + ρ2 sin2 ϕ = 4ρ cosρ, |
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ρ = 4cos ϕ . |
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Straight lines |
y = 0 and y = x in the polar components have the form ϕ = 0 |
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correspondently. Hence, polar components ϕ and ρ of the points which |
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S = ∫∫ dxdy = ∫∫ρdρdϕ = ∫4 dϕ ∫ |
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Example 9. Find the volume V bounded by a parabolic cylinder y = x2 and planes z = 0, z = 2 – y.
Solution. Examine D. Let’s consider a projection of the given figure on the xy-plane. We obtain:
− 2 ≤ x ≤ 2
and x2 ≤ y ≤ 2.
From Fig. 5.19 it is clear that the figure bounded from above by the plane z = 2 – y. Therefore,
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V = ∫∫(2 − y)dxdy = ∫ |
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Fig. 5.19
117
Example 10. Find the volume V bounded by surfaces z = 4 − 2y2 , z = 0, x = 0, y = 0 and x + 2 y = 2.
Solution. The given body is limited from above by parabolic cylinder z = 4 − 2y2 , from below is bounded by a coordinate plane Оху, from sides are bounded by planes Охz, Оyz and x + 2 y = 2 (Fig. 5.20).
z
4
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x
Fig. 5.20
Therefore, the cylindrical body is given. Domain of an integration D is a triangle: 0 ≤ y ≤ 1, 0 ≤ x ≤ 2 − 2y. We receive
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Example 11. Find the surface area of the part of a paraboloid of rotation 2z = x2 + y2 (x ≥ 0), between planes x = 0 and z = 8.
Solution. Half-circle of the |
radius R = 4 |
with the centre in the origin: |
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the given surface on the xy-plane. We |
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Then by the formula (5.32) we have
Sσ = ∫∫ 1+ x2 + y2 dxdy.
Dxy
Let’s conduct calculations in a polar system:
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Sσ = ∫∫ 1+ x2 + y2 dxdy = ∫ dϕ∫ |
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Example 12. Find the mass m of D: y = 0, x + y = 2, y = x2 and γ = y2 x.
Solution. Let’s find the mass of the figure (Fig. 5.21) bounded by two
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y = 0, x + y = 2 and a parabola y = x2 . The point of intersection |
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119
Example 13. Find coordinates of the center of mass (xc, yc) for homogeneous
plate, γ = 1, if D: y
y
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−π |
≤ x ≤ |
π |
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2 |
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Solution. The given plate is symmetrical with respect to Oy, then xc = 0 (Fig. 5.22).
Let’s find the static moment of inertia and mass. With the help of the formula (5.35) we find the static moment of inertia of the planar figure
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Fig. 5.22 |
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Now we calculate mass of the figure |
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m = ∫∫ dxdy = ∫ dx |
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0cos x dx = ∫ cos xdx = sin x |
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Micromodule 5
CLASS AND HOME ASSIGMENTS
To place limits of an integration in a double integral ∫∫ f (x, y)dxdy over the
domain D if:
D
1.D is a triangle АВС, where A(0; 0) , B(4;1) , C(4; 4) .
2.D is a tetragon АВСD, where A(−1; 1) , B(−1; 4), C(1; 6) and D(4; 6).
120