Calculate the mass of a lamina that occupies the

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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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