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Unformatted text preview: with respect to the y axis] 5. Calculate the mass of a lamina that occupies the plane region R bounded by the curve
1
(x − 1)2 + y 2 = 1 with density function ρ(x, y ) =
.
2 + y2
x
1
1
1
becomes ρ(r, θ) = √ = .
Clearly, this problem is best represented in polar coordinates. First, ρ(x, y ) =
2
2 + y2
r
r
x
Next, the region in the plane is given by: x2 − 2x + 1 + y 2 = 1, or r2 − 2r cos θ = 0. Thus we have r2 = 2r cos θ, or
r = 2 cos θ.
π /2
π /2 cos θ
1
2 cos θdθ
r dr dθ = 2
Therefore, using symmetry, we have mass= 2
r
0
0
0
π
2 = 4 sin θ = 4(1) − 0 = 4 0 2/ sin θ 3π/4 r dr dθ. 6. Sketch the region of integration for
1 π/4 π
3π
2
Notice that we have
≤θ≤
. Also, r = 1 is the circle of radius 1 centered at the origin, and if r =
, then
4
4
sin θ
r sin θ = 2 or y = 2.
Therefore the region of integration is as follows:
y θ = π/4 θ = 3π/4 y=2 2 R
r=1
−1 −2 x 1 2 −1
−2 7. Find the surface area of the surface S where S is ﬁrst octant portion of the hyperbolic paraboloid z = x2 − y 2 that is
inside the cylinder x2 + y 2 = 1.
R 2
2
fx + fy + 1 dA. Here, z = f (x, y ) = x2 − y 2 , so fx = 2x and fy = −2x R (2x)2 + (−2y )2 + 1 dA = Recall that S =
So we have S = in the ﬁrst octant (so x ≥ 0, y ≥ 0, and z ≥ 0).
1 π
2 π
2 4r2 + 1r dr dθ = Converting to polar coordinates gives:
0
π
2 1
=
12 0 4x2 + 4y 2 + 1 dA where R is portion of the cylinder x2 + y 2 = 1
R 0 0 3
21 2
· (4r + 1) 2
38 dθ
0 3
3
1π
π
52 − 1 =
52 − 1 .
5 − 1 dθ =
12 2
24
3
2 2 4− y 4 8. Let I =
0 x + yz dz dy dx. Sketch the solid Q over which the iterated integral takes place, and rewrite the x2 0 iterated integral in the order dx dz dy . DO NOT EVALUATE THE INTEGRAL.
z 4 y
z=4−y 1 4 2 y = x2 x Notice that we have 0 ≤ z ≤ 4 − y , x2 ≤ y ≤ 4, and 0 ≤ x ≤ 2
√
Then we also have 0 ≤ x ≤ y , 0 ≤ z ≤ 4 − y , and 0 ≤ y ≤ 4
Thus the integral in the required order is:
0 √ 4− y 4 0 y x + yz dx dz dy
0 9. Find the mass of the cylinder of radius 3 between z = 0 and z = 4 if the density at the point (x, y, z ) is given by
δ (x, y, z ) = z + x2 + y 2 .
This mass integral is best represented in cylindrical coordinates:
4 3 2π 4 3 2π zr + r2 dz dr dθ (z + r)r...
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 Fall '11
 James
 Math, Vector Calculus

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