Higher_Mathematics_Part_3
.pdfExample 2. Evaluate the surface integral of the first type I = ∫∫ z(x + 2y)dσ,
σ
where σ is the part of the surface z = 1− x2 bounded by the planes y = 0 and y = 3 (Fig. 8.9).
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Solution. The projection of the given surface to the plane Оху is the |
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rectangular: |
−1 ≤ x ≤ 1, |
0 ≤ y ≤ 3 (Fig. |
8.10). We’ll find |
partial derivatives |
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z′ |
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Then |
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d |
σ = |
1+ (z′ )2 + (z′ )2 dxdy = |
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dxdy = |
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Now we’ll evaluate the surface integral |
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I = ∫∫ z(x + 2y)dσ = I = ∫∫ |
1− x2 (x + 2y) |
dxdy |
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= ∫∫ (x + 2y)dxdy = |
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= ∫ dx∫ (x + 2y)dy = ∫ (xy + y |
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dx = ∫ (3x + 9)dx |
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Fig. 8.8 |
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Fig. 8.9 |
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Fig. 8.10 |
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Example 3. Find |
coordinates |
of the centre of mass |
for the |
hemisphere |
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z = R2 − x2 − y2 |
(Fig. 8.11) if its surface density at each point is numerically |
equal to the distance of this point to the radius, perpendicular to the basis of the hemisphere.
Solution. Under the condition of the problem the surface density at a point
(x, y, z) is defined by the formula γ = x2 + y2 . From symmetry of the hemisphere to the axis Оz and the function γ(x, y) comparatively to the point (0; 0) it follows, that the centre of mass is placed on the axis Оz. So, xc = yc = 0 and we’ll define coordinate zc accordingly the formula (8.7).
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Let’s transform an element dσ. As
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R2 − x2 − y2 |
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R2 − x2 − y2 |
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then |
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1 |
+ (z′ )2 |
+ (z′ )2 |
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x2 |
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y2 |
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Rdxdy |
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dσ = |
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R2 − x2 − y2
Considering that the projection of the surface to the plane Оху is the disk of
radius R, bounded with the circle x2 + y2 |
= R2 , |
we make calculations in polar |
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coordinates system. We have |
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x2 + y2 dxdy |
2π |
R |
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ρ |
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R |
ρ2dρ |
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m = R ∫∫ |
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= R ∫ dϕ∫ |
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ρdρ = 2πR∫ |
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ρ = R sin t, |
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= 2πR∫2 |
R2 sin2 t R costdt |
= 2πR3 ∫2 sin2 tdt = |
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dρ = R costdt, |
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0 ≤ t ≤ π / 2 |
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π |
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sin 2t |
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∫(1− cos 2t)dt = πR |
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t − |
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2π |
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∫∫ zγ dσ =R ∫∫ |
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x2 + y2 dxdy =R ∫ dϕ∫ρ2dρ = |
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So, |
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zc = |
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Example 4. Evaluate the surface integral of the second type |
I = ∫∫ xdydz + |
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+ zdxdz + 3dxdy, |
where |
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σ |
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the plane |
2x − 3y + 3z − 6 = 0 ( x ≥ 0, y ≤ 0, z ≥ 0 ).
Solution. The given surface σ, being the part of plane, is represented on the Fig. 8.12. The normal n corresponding to the upper side of the surface forms the
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acute angles with axes Oh and Oz and the obtuse angle with the axis Оу. Actually, the normal n = {2; − 3; 3} has such directing cosines:
cos α = 2 |
> 0, |
cosβ = |
−3 |
< 0, cos γ = |
3 > 0 . |
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z |
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R |
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R у |
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Fig. 8.11 |
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Fig. 8.12 |
Therefore the surface integral is reduced to the sum of three double integrals over the domains represented on Fig. 8.13, the first and third taken with the sign «+» and the second taken with the sign «–».
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у |
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Dyz |
2 |
2 |
Dxz |
О |
3 х |
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–2 |
О у |
О |
3 х |
–2 |
Dxy |
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Fig. 8.13
So, we have
I = ∫∫ xdydz + zdxdz + 3dxdy = ∫∫ xdydz − ∫∫ zdxdz + ∫∫ 3dxdy =
σ |
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Dyz |
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y+ 2 |
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3 |
6− 2x |
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6 + |
3y − 3z |
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= ∫ dy ∫ |
dz −∫ dx |
∫ zdz +3 |
3 2 = |
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= ∫ |
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dy − |
∫ |
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dx + 9 = −4 − 2 + 9 = 3. |
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Example 5. Evaluate the surface integral of the second type ∫∫ zdxdy, if σ is
σ
the external side of sphere x2 + y2 + z2 = R2 (Fig. 8.14).
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z G
R n
О R у
G
R n
х
Fig. 8.14
Solution. The upper and the lower hemispheres are projected on the plane Оху in same domain, i.e.
the circle limited by a circle x2 + y2 = R2 . That is why we divide the surface σ into the part σ1 and the part
σ2 , where σ1 is the upper hemisphere z = R2 − x2 − y2 ,
and σ2 |
is the lower hemisphere z = − R2 − x2 − y2 . |
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So, |
∫∫ zdxdy = ∫∫ zdxdy + ∫∫ zdxdy . |
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σ |
σ1 |
σ2 |
We’ll reduce each of integrals to the double one, considering, that a normal vector on the selected side of the upper hemisphere forms with the axis Oz an acute angle, and an obtuse angle with bottom hemisphere. So,
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∫∫ zdxdy = + ∫∫ |
R2 − x2 − y2 dxdy , |
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σ1 |
Dxy |
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∫∫ zdxdy = − ∫∫ − |
R2 − x2 − y2 dxdy = ∫∫ R2 − x2 − y2 dxdy . |
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σ2 |
Dxy |
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Dxy |
Then |
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∫∫ zdxdy = 2 ∫∫ |
R2 − x2 − y2 dxdy = I. |
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σ |
Dxy |
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Now we’ll pass to polar coordinates and we’ll get
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2π |
R |
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I = ∫∫ zdxdy = 2 ∫ dϕ∫ |
R2 − ρ2 ρdρ = |
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σ |
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0 |
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R |
d (R |
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− ρ |
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= 4π∫ R2 − ρ2 |
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= − |
π (R2 − ρ2 )3 |
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πR3. |
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Example 6. Evaluate the surface integral of the second type. ∫∫ 2xdydz − ydxdz,
σ
where σ is the external side of the part of the surface of the cylinder x2 + y2 = 1,
0 ≤ z ≤ 1, x ≥ 0, y ≥ 0 (Fig. 8.15). |
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Solution. We’ll consider the given integral |
as the sum of two |
integrals |
I = I1 + I2 . For evaluation of the surface integral |
I1 = ∫∫ 2xdydz we project the |
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σ |
0 ≤ z ≤ 1 |
surface σ on the plane Oyz. We’ll get the rectangular σ yz : 0 ≤ y ≤ 1, |
(Fig. 8.16). Now we’ll solve the equation of the cylinder comparatively to the x: x = 1− y2 (the condition x ≥ 0 ). As the normal vector n at any point of the
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selected side of the surface σ forms with the axis Ox an acute angle, then we take the corresponding double integral with the sign «+». We have
I1 = ∫∫ 2xdydz = ∫∫ 2 |
1 |
1 |
1 |
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1− y2 dydz = 2∫ |
1− y2 dy∫ dz = 2∫ 1− y2 dy = |
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σ |
σ yz |
0 |
0 |
0 |
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y = sin t, |
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π |
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sin 2t |
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= |
dy = cos tdt, |
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= ∫cos |
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tdt = |
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∫(1+ cos 2t)dt = |
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t |
+ |
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= |
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0 ≤ t ≤ π |
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Fig. 8.15 |
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Fig. 8.16 |
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Similarly we evaluate the integral I2 = −∫∫ ydxdz. For this purpose we
σ
project the surface on the plane Оxz and solve the equation of a surface with
respect to y: y = 1− x2 . Then we pass to the double integral (a sign of double integral is the same, as the sign of the surface integral). Then
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1 |
1 |
π , |
I2 = −∫∫ ydxdz = −∫ 1− x2 dx∫ dz = − |
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σ |
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4 |
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And finally we receive I = π − |
π = |
π . |
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4 |
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Example 7. Evaluate the surface integral of the second type |
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I = ∫∫ x2dxdz + xdxdz + xzdxdy, |
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σ |
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where σ is the upper side of that surface part |
y = x2 + z2 |
which is placed in the |
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first octant between planes y = 0 and |
y = 1. |
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Solution. We’ll represent the surface σ. The equation y = x2 + z2 determines a paraboloid of the rotation around the axis Oy. And its part which is placed in the first octant, crosses coordinate plane Oyz on the parabola y = z2 , and plane Oxy
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1 x
z |
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z = 1−x2 |
on the parabola |
y = x . With the plane |
y = 1 the |
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y = z2 |
paraboloid is crossed along the circle x2 + z2 = 1 , |
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σ |
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n |
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the quarter of this circle lays in the first octant. |
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Eventually, if |
y = 0, then x2 + z2 = 0. |
Only one |
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point (the Origine) satisfies this equation. As a result |
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of the analysis carried we constract the surface σ |
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(Fig. 8.17). |
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n at any |
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y = x2 |
Let’s notice, that the normal vector |
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point of the selected side of a surface |
σ creates |
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Fig. 8.17 |
acute angle with axes Ox and Oz, and it creates |
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obtuse angle with the axis Oy. |
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Let’s evaluate three integrals in series. |
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1) |
I1 = ∫∫ x2 dydz . |
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σ |
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From the surface equation y = x2 + z2 it followes that x2 = y − z2 and we pass to double integral behind a projection σ yz . As the normal vector n creates acute angle with the axis Ox (see Fig. 6.16) before double integral we put a plus sign:
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I1 = ∫∫ ( y − z |
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= ∫ dy ∫ ( y − z |
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= ∫ yz − |
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dy = |
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2) I2 = ∫∫ xdxdz.
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The integrand does not contain variable y, therefore we pass directly to the double integral behind a projection σxz . From Fig. 8.17 it is сlear, that a normal
n corresponding to the upper side of a surface σ , creates an obtuse angle with the axis Oy, therefore double integral we take with a minus sign:
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1− x2 |
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I2 |
= − ∫∫ xdxdz = −∫ xdx |
∫ |
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∫ 1− x2 d (1− x2 ) = |
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3)I3 = ∫∫ xzdxdy .
σxy
From the surface equation we find that z = ± y − x2 , but we take a plus sign before the root because z ≥ 0 in the first octant, and we pass to the double
196
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integral behind a projection σxy , having taken integral with a plus sign (the normal n creates an acute angle with the axis Oz):
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I3 = ∫∫ x |
y − x2 dxdy = ∫ dy ∫ x |
y − x2 dx = − |
∫ dy ∫ |
y − x2 d( y − x2 ) = |
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∫ |
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∫ y3 dy = |
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Example 8. Evaluate the surface integral of the second type |
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I = ∫∫ xdydz + zdxdz + 3dxdy, |
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where σ is the external side of the pyramid bounded by the |
x = 0, |
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and 2x − 3y + 3z − 6 = 0 (Fig. 8.12). |
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Solution. As the surface is closed, we apply Ostrogradsky-Gauss formula. We’ll have
P = x, Q = z, R = 3; |
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= 1, |
∂Q |
= 0, |
∂R |
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I = ∫∫∫(1+ 0 + 0)dxdydz =Vpymid |
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G |
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Example 9. Evaluate the surface integral of the second type
I = ∫∫(x + 3z)dydz − 2ydxdz + (z − y)dxdy,
σ
where σ is the external side of the cone part x2 + y2 = z2 placed between planes z = 0 and z = 1 (Fig. 8.18).
Solution. It is impossible to apply OstrogradskyGaussian formula to the given integral directly. On the other hand, evaluation of integral by means of designing a surface on coordinate planes leads to rather bulky calculations. We’ll apply the following method. Let’s close the surface with the circle σ1 (a part of the plane
z = 1 placed inside a cone). Then
I = I1 − I2 ,
z
n
1 σ2
σ1
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Fig. 8.18 |
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where
I1 = ∫∫ (x + 3z)dydz − 2 ydxdz + (z − y)dxdy,
σ+σ1
I2 = ∫∫(x + 3z)dydz − 2ydxdz + (z − y)dxdy.
σ1 |
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As |
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P = x + 3z, Q = −2 y, R = z − y; |
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= 1− 2 + 1 = 0, |
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then I1 = 0 and consequently I = −I2 , i.e. the required integral is reduced to the surface l integral over the disk σ1 . The projections of this disk on the coordinate planes Oxz and Oyz are segments. So,
I = −I2 = − ∫∫ (z − y)dxdy = − ∫∫ (1− y)dxdy.
σ1 Dxy
The sign of the double integral is not changed, as the angle between the normal vector n and the axis Oz is equal to zero ( cos γ = 1 > 0 ).
Having passed to polar coordinates in double integral, we get
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I = − ∫ dϕ∫(1− ρ cos ϕ)ρdρ = − ∫ |
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Micromodule 8 |
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CLASS AND HOME ASSINMENTS |
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Evaluate the surface integral of the first type over the given surface. |
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1. |
∫∫(x2 + y2 + z2 )dσ, |
if σ is the hemisphere |
z = |
R2 − x2 − y2 . |
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∫∫(x2 + 3y2 + z2 + 5)dσ, |
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y = 0 |
σ |
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3. |
∫∫ (x2 + ( y2 + z2 )2 )dσ, |
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x + y + z = 2 , which |
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has been cut out by the cylinder |
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4. |
∫∫(2x + 3y + 5z)dσ, if σ is the part of the plane 2x + 3y + 5z = 30, placed |
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in the first octant. |
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Evaluate the surface integral of the second type over the given surface. |
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5. |
∫∫ ( y2 + z2 )dxdy, if σ is the external side of the cylinder part z = |
9 − x2 , |
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placed between planes y = 0 |
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6. |
∫∫ (x2 + y2 + z2 )dxdy, |
if |
σ is the external side |
of a |
hemisphere |
part |
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y = |
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y = |
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7. |
∫∫ x2 dydz + y2dxdz + z2dxdy, |
if |
σ is the external |
side |
of a sphere |
part, |
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8. |
∫∫ zdxdy − ydydz, if σ is |
triangle formed by section of a plane |
6x − |
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− 3y + 2z = 6 with coordinate planes and a normal to the selected side creates an |
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acute angle with the axis Oz. |
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9. |
∫∫ xdydz + ydxdz + zdxdy, |
if σ is the external side of a sphere x2 + y2 + |
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+ z2 = R2 . |
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10. Calculate coordinates |
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mass centre |
of a |
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2z = 4 − x2 − y2 (z ≥ 0) . |
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Answers |
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1. |
2πR4. 2. 52 2π. 3. 29 |
3π / 6. |
4. |
450 14. |
5. 88. |
6. |
π / 2. 7. 3πR4 /8. 8. 3. |
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9. 4πR3. 10. (0; 0; 307 − 15 5)/310. |
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Micromodule 8
SELF—TEST ASSINMENTS
8.1. Evaluate the surface integrals of the first type over the surface G (Table 8.1).
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Table 8.1 |
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№ |
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Integral |
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The equation |
Additional conditions |
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of surface G |
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8.1.1 |
∫∫(x |
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)dσ |
z = x2 + y2 |
Surface G is bounded by |
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planes z = 0 , z = 2 |
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G |
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199 |
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Continuity of table 8.1 |
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№ |
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Integral |
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The equation |
Additional conditions |
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of surface G |
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8.1.2 |
∫∫ ydq |
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z = |
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9 − x2 − y2 |
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G |
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8.1.3 |
∫∫ xyzdq |
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x + y + z = 1 |
Surface G is placed in the first |
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G |
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octant |
8.1.4 |
∫∫(3z + 6x + 4y)dq |
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Surface G is placed in the first |
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2 |
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octant |
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G |
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3 |
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8.1.5 |
∫∫ |
16 − x2 − y2 dq |
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z = − 16 − x2 − y2 |
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8.1.6 |
∫∫(x2 + y2 )dq |
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x2 + y2 + z2 = 4 |
z ≥ 0 |
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8.1.7 |
∫∫ x2 z2dq |
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y = |
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8.1.8 |
∫∫ |
y |
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dq |
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z |
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− x2 = 0 |
Surface G is bounded by planes |
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x = 0 , x = 1 |
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16 |
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9 |
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8.1.9 |
∫∫ |
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dq |
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x2 + z2 = 16 |
Surface G is bounded by planes |
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x2 + y2 |
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y = 0, y = 2 |
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G |
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8.1.10 |
∫∫(x2 + z2 )dq |
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y = 1− x2 − z2 |
y ≥ 0 |
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8.1.11 |
∫∫ |
(x |
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2 |
+ z |
2 |
)dq |
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y = |
x2 + z2 |
Surface G is bounded by planes |
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y = 0, y = 4 |
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8.1.12 |
∫∫ xdq |
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z = |
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16 − x2 − y2 |
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G |
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8.1.13 |
∫∫ y2 zdq |
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x + 2y + z = 1 |
Surface G is placed in the first |
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G |
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octant |
8.1.14 |
∫∫(z + 2x + 3y)dq |
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Surface G is placed in the first |
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3 |
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octant |
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G |
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2 |
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8.1.15 |
∫∫ |
25 − x2 − z2 dq |
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y = − |
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25 − x2 − z2 |
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G |
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8.1.16 |
∫∫( y2 + z2 )dq |
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x2 + y2 + z2 = 36 |
x ≥ 0 |
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G |
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200
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