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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y |
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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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1− x2 |
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σ |
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= ∫ dx∫ (x + 2y)dy = ∫ (xy + y |
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dx = ∫ (3x + 9)dx |
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= 18. |
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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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then |
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1 |
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x2 |
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y2 |
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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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sin 2t |
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= πR |
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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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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 = |
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< 0, cos γ = |
3 > 0 . |
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R |
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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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2 |
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Dxz |
О |
3 х |
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–2 |
О у |
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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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6− 2x |
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3y − 3z |
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dz −∫ dx |
∫ zdz +3 |
3 2 = |
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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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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 |
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1− y2 dydz = 2∫ |
1− y2 dy∫ dz = 2∫ 1− y2 dy = |
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σ |
σ yz |
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y = sin t, |
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π |
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dy = cos tdt, |
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tdt = |
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∫(1+ cos 2t)dt = |
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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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π , |
I2 = −∫∫ ydxdz = −∫ 1− x2 dx∫ dz = − |
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And finally we receive I = π − |
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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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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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− ρ |
cos ϕ |
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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 = |
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2. |
∫∫(x2 + 3y2 + z2 + 5)dσ, |
if σ is the part of a cone placed between planes |
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y = 0 |
σ |
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and y = 2. |
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3. |
∫∫ (x2 + ( y2 + z2 )2 )dσ, |
if σ is the part of the plane |
x + y + z = 2 , which |
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y2 + z2 |
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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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7. |
∫∫ x2 dydz + y2dxdz + z2dxdy, |
if |
σ is the external |
side |
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part, |
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placed in first octant. |
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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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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 |
of |
mass centre |
of a |
homogeneous |
surface |
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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. |
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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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z = x2 + y2 |
Surface G is bounded by |
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planes z = 0 , z = 2 |
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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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8.1.3 |
∫∫ xyzdq |
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x + y + z = 1 |
Surface G is placed in the first |
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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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octant |
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8.1.5 |
∫∫ |
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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 |
∫∫ |
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+ z |
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dq |
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Surface G is bounded by planes |
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x = 0 , x = 1 |
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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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8.1.10 |
∫∫(x2 + z2 )dq |
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y = 1− x2 − z2 |
y ≥ 0 |
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8.1.11 |
∫∫ |
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+ z |
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)dq |
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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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8.1.13 |
∫∫ y2 zdq |
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x + 2y + z = 1 |
Surface G is placed in the first |
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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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octant |
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8.1.15 |
∫∫ |
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8.1.16 |
∫∫( y2 + z2 )dq |
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x2 + y2 + z2 = 36 |
x ≥ 0 |
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