Unformatted text preview: Math 2220 Prelim II Solutions Spring 2010 Note: Each problem is worth 14 points except numbers 5 and 6 which are 15 points. x2 dA where D is the region in the second quadrant between the 1. Compute 2 2 D x +y circles x2 + y 2 = 1 and x2 + y 2 = 4 . Solution. In polar coordinates the region D is given by the inequalities 1 ≤ r ≤ 2 and π/2 ≤ θ ≤ π . The integral then becomes
π π/2 1 2 r 2 cos2 θ r dr dθ = r2 2 π r dr
1 π/2 cos θ dθ = = 3 2 2 r2 2 2 1 π π/2 1 + cos 2θ dθ 2
π π/2 θ sin 2θ + 2 4 = 3 2 π 4 = 3π 8 2. Let D be the region in the ﬁrst quadrant bounded by the curves y = xα and y = xβ for 0 < β < α . Compute
D xy dA in two ways, integrating in the order dx dy and in the order dy dx , and verify that the answer is the same in both cases. Solution. The graphs of xα and xβ intersect at the points (0, 0) and (1, 1) . Between these two points the graph of xβ is above the graph of xα . Integrating in the order dy dx we then have
1 0 xβ 1 xy dy dx =
xα 0 12 xy 2 xβ dx =
xα 1 2 1 0 x2β +1 − x2α+1 dx = 1 2 x2β +2 x2α+2 − 2β + 2 2α + 2 1 0 = = 1 2 1 1 − 2β + 2 2α + 2 α−β 4(α + 1)(β + 1) If we change to the order dx dy then the integral becomes
1 0 y 1/α 1 xy dx dy =
y 1/β 0 12 xy 2 y 1/α y 1/β dy = 1 2 1 0 y (2/α)+1 − y (2/β )+1 dy
1 1 = 2 y (2/β )+2 y (2/α)+2 − (2/α) + 2 (2/β ) + 2 = 1 2 =
0 1 2 1 1 − (2/α) + 2 (2/β ) + 2 = α−β 4(α + 1)(β + 1) α β − 2α + 2 2β + 2 So we get the same answer in both cases. Math 2220
2 Prelim II Solutions
2−x 0 0 4−2y Spring 2010 3. Consider an integral
0 f (x, y, z ) dz dy dx . (a) Sketch the region of integration. Solution. We can determine the region from the limits of integration. For the dz integration we see that z goes from z = 0 right). For the dy and dx integrations we are looking at the shadow of the region in the xy plane, a triangle in the ﬁrst quadrant cut oﬀ by the line y = 2 − x . The 3 dimensional shape shown in the ﬁgure, sort of a pyramid lying on its side. since we have not speciﬁed what the function f (x, y, z ) is). Solution. For the order dx dy dz we are projecting to the shadow in the yz plane, which is the triangle in the ﬁrst quadrant of the yz plane cut oﬀ by the line z = 4 − 2y , or to x = 2 − y (a vertical plane). The integral is then
4 0 0 2−(z/2) 0 2−y (the xy plane) to z = 4 − 2y (a plane sloping down to the region of integration lies above this triangle and below the plane z = 4 − 2y , so it has the (b) Set up the integral in the order dx dy dz (but do not attempt to evaluate the integral y = 2 − (z/2) . For the initial dx integration we have x going from x = 0 (the yz plane) f (x, y, z ) dx dy dz. 4. Compute the volume of the region bounded by the xy plane and the two surfaces several symmetric pieces of equal volume. z = 1 − x2 and z = 1 − y 2 (these are parabolic cylinders). Hint: break the region up into Solution. The region is shown in the ﬁgure on the left below. It is tentshaped, and the projection of the region to the xy plane is the square −1 ≤ x ≤ 1 and −1 ≤ y ≤ 1 . The region is symmetric under reﬂection across the yz plane and the xz plane. It is also symmetric under reﬂection across the vertical planes x = ±y . We can compute the volume Math 2220 Prelim II Solutions Spring 2010 by computing the volume of the piece shown in the right half of the ﬁgure and multiplying this by 8 . For this piece the curved upper surface is part of the surface z = 1 − x2 , so the volume of this piece is
1 0 0 x 0 1−x2 1 x 1−x2 0 2 x 0 1 x 0 1 dz dy dx =
0 1 0 z dy dx =
0 1 (1 − x2 ) dy dx
3 =
0 (1 − x )y dx =
0 1 4 (x − x ) dx = x4 x2 − 2 4 1 =
0 1 4 Therefore the volume of the region is equal to 8 × For example:
1 = 2. There are other ways to compute the volume, using diﬀerent orders for dx , dy , and dz .
√ 1−z 0 0 √ 1−z 1−y 2 0 0 √ 1 4
0 dx dy dz or 4
0 1−z dx dz dy 5. Find the center of gravity of the constantdensity solid cone consisting of all points (x, y, z ) satisfying the inequalities x2 + y 2 ≤ z ≤ a , where a is an arbitrary constant. Solution. We use cylindrical coordinates. The inequalities The volume of the cone is given by the integral
2π a 0 r a 2π a x2 + y 2 ≤ z ≤ a then become r ≤ z ≤ a . 1 dV =
D 0 r dz dr dθ =
0 2π 0 rz
2 a r 2π a 0 dr dθ =
0 3 a 0 (ar − r 2 ) dr dθ
2π =
0 r ar − 2 3 dθ =
0 a3 πa3 dθ = 6 3 Using the rotational symmetry of the cone we can see that the center of gravity lies on the z axis so its x and y coordinates are 0 . In order to compute the z coordinate we need the integral
2π a 0 r a 2π a 0 2π 0 z dV =
D 0 r z dz dr dθ =
0 rz 2 2 a r 1 dr dθ = 2
4 a 0 2π 0 0 a (a2 r − r 3 ) dr dθ
2π 0 = 1 2 ar r − 2 4 22 dθ = 1 2 a4 πa4 dθ = 4 4 Finally, the z coordinate of the center of mass is z= ¯ z dV πa4 /4 3a = = 3 /3 πa 4 1 dV D
D Math 2220 Prelim II Solutions Spring 2010 6. Find the volume of the region that lies inside the sphere x2 + y 2 + z 2 = 2z and outside the sphere x2 + y 2 + z 2 = 2 . Solution. The sphere x2 + y 2 + z 2 = 2 has radius √ 2 with center at the origin. The equation for the other sphere can be 1 and center at (0, 0, 1) . In cross section the two spheres look like the circles in the ﬁgure at the right. The region we want is the shaded region. We use spherical coordinates to compute the volume. In spherical coordinates the lower √ sphere has equation ρ = 2 and the upper sphere has equation ρ2 = 2z = 2ρ cos ϕ , or ρ = 2 cos ϕ . These two spheres intersect in the circle where the plane z = 1 intersects the two spheres, so ϕ = π/4 on this circle. Therefore the region can be described in spherical coordinates as √ 2 ≤ ρ ≤ 2 cos ϕ 0 ≤ ϕ ≤ π/4 0 ≤ θ ≤ 2π rewritten as x2 + y 2 + (z − 1)2 = 1 so this sphere has radius The volume of the region is given by the integral
2π 0 0 π /4 2 cos ϕ √ 2 ρ sin ϕ dρ dϕ dθ =
0 0 2π 2 2π π /4 ρ3 sin ϕ 3 2 cos ϕ √ 2 dϕ dθ π /4 √ 1 = dθ (8 cos3 ϕ sin ϕ − 2 2 sin ϕ) dϕ 30 0 π /4 √ √ π 2π = = −2 cos4 ϕ + 2 2 cos ϕ 7−4 2 3 3 0 Here we used the substitution u = cos ϕ to compute the dϕ integral. 7. Let D be the region in R2 deﬁned by the inequalities y ≤ x ≤ 2y and 3 ≤ x + y ≤ 4 . 1 Compute dx dy by using a change of variables to convert D into a square. Hint: D xy Rewrite the inequalities y ≤ x ≤ 2y as 1 ≤ x/y ≤ 2 . and 1 ≤ x/y ≤ 2 , and the easiest way to do this is to choose new variables u = x/y and Solution. We want to simplify the region D deﬁned by the inequalities 3 ≤ x + y ≤ 4 v = x + y so that the inequalities become 1 ≤ u ≤ 2 and 3 ≤ v ≤ 4 , which deﬁne a square. To carry out the change of variables from x and y to u and v we will need to express x and y in terms of u, v , i.e., we need to solve for x and y in the system x/y = u x + y = v. Math 2220 Prelim II Solutions Spring 2010 Solving the ﬁrst equation for x gives x = uy and substituting this in the second equation gives uy + y = v , which can be solved for y to get y = x = uy gives x =
uv u+1 . Thus we have x = uv u+1 and y = v u+1 . v . u+1 Then the earlier equation Next we need to compute the Jacobian determinant: ∂ (x, y ) = det ∂ (u, v )
∂x ∂u ∂y ∂u ∂x ∂v ∂y ∂v = det v (u+1)2 v − (u+1)2 u u+1 1 u+1 = v v + uv = . 3 (u + 1) (u + 1)2 The change of variables formula gives 1 dA = xy 1
D′ v uv u+1 u+1 D D′ 1 ∂ (x, y ) dA = x(u, v )y (u, v ) ∂ (u, v ) 1 dA, uv v dA = (u + 1)2 D′ where D′ is the region in the uv plane corresponding to D , so D′ is the square 1 ≤ u ≤ 2 , 3 ≤ v ≤ 4 . Thus we get
2 1 3 4 1 dv du = uv 2 1 1 du u 4 3 1 dv = v ln u 2 1 ln v 4 3 = ln 2(ln 4 − ln 3) ...
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This note was uploaded on 04/21/2010 for the course MATH 2220 taught by Professor Parkinson during the Spring '08 term at Cornell.
 Spring '08
 PARKINSON
 Math, Multivariable Calculus

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