Calculus III. Math 233 Spring In-term exam April 11th. Suggested solutions

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1 Calculus III Math Spring 7 In-term exam April th. Suggested solutions This exam contains sixteen problems numbered through 6. Problems 5 are multiple choice problems, which each count 5% of your total score. Problem 6 will be hand-graded and counts 5% of your total score. Problem Evaluate y sin(πx) dx dy. D) 5 π E) F) π G) H) 7π A) π B) π C) π This is a straightforward calculation, y sin(πx) dx dy π [ π ] x y cos(πx) dy y dy π x [ y ] 5 π. ( π y ) (( ) ) dy

2 Problem Use Lagrange multipliers to find the maximum value of the function A) B) G) H) f(x, y, z) x y + z on the sphere (x ) + (y ) + (z ). C) D) + E) F) 6 Let g(x, y, z) (x ) + (y ) + (z ). Then the method of Lagrange multipliers says that a maximum can only occur at a point where f λ g. Because f,, and g (x ), (y ), (z ), we need to solve λ(x ) λ(y ) λ(z ) (x ) + (y ) + (z ). The first three equations tell us that x (y ) z. By the last equation, (x ), or x ±. Because y x and z x, the value of f is f(x, y, z) x y + z x ( x) + x ( x ). The maximum value is therefore attained when x, where it is.

3 Problem Find the surface area of the part of z x y that lies above the xy-plane. A) 5π B) π C) D) + 6 π π E) 5 π F) (5 5 )π 6 G) π 7 H) This is about the surface area of a graph, so we can use formula 6 on page 87 of our calculus book. We need the partial derivatives of z: z x x, z y y. The function we need to integrate, is + ( x) + ( y) + (x + y ). Then we need to find the domain: since z x y, we get x + y. So it looks like it would be best to use polar coordinates. Now we can set up the integral, π + r r dr dθ This calls for a substitution. Defining u + r, we get du 8r dr, so r dr 8 du. Hence, π + r r dr dθ π π 5 π 8 u du dθ [ u/] u5 dθ 8 u ( 5 5 ) dθ 6( 5 5 ) π.

4 Problem Find the volume of the solid bounded by z x y and the xy-plane. A) B) C) D) E) π F) π G) π H) π To find the volume we will integrate the height of the solid, over the projection of the solid in the xy-plane. The projection of the solid is bounded by the circle x + y, while the height is x y. Everything is conveniently described using polar coordinates, so we calculate V π π ( r ) r dr dθ [ r r] dθ π π dθ π. r r dr dθ

5 Problem 5 Let D be the triangle with vertices (, ), (, ), and (, ). Calculate xy da. D A) 6 B) 7 C) 57 D) 5 E) F) G) 7 H) First, let s draw a picture of our triangle D: This can be seen as either a Type I or a Type II region. As a Type I region, it produces the following integral, [ xy da xy dy dx xy] y dx x( ( x) ) dx y x D x x( x x ) dx x x dx [ x 8 x] Problem 6 Find the Jacobian (x,y) (u,v) of the transformation x u + v, y u v. A) B) 5 C) D) E) F) G) H) 6 The Jacobian is given by (x, y) (u, v) x y u v x y v u ( ). 5

6 Problem 7 Compute the volume of the pyramid with a rectangular base that is bounded by the five planes x + z, x + z, y + z, y + z and z. Hint: Use symmetry to calculate the volume of the pyramid as four times the volume of a tetrahedron. A) 6 B) 8 C) 8 D) 7 E) 6 F) 56 G) H) If we project the pyramid onto the xy-plane, we get the following: For instance, the plane x + z intersects the xy-plane in the line x (we find this by substituting z in the equation for the plane). We can calculate the total volume as four times the volume of one of the tetrahedrons of which the pyramids consists, for example the one with vertices (,, ), (,, ), (,, ) and (,, ). This one is in fact the solid between D and the plane x + z, where D is the triangle in the xy-plane with vertices (, ), (, ) and (, ). This region D is a Type I region, so we can set up the integral to compute the volume: D ( x) da x x ( x) dy dx ( x) x dx So the volume of the pyramid is [ ( x)y ] yx y x dx x 6x dx [ x x ] 6

7 Problem 8 Evaluate E dv, where E is the solid bounded by two spheres: x +y +z E {(x, y, z): x + y + z }. A) 8π H) π B) π C) π D) π E) π F) π G) π In spherical coordinates E is described by ρ, φ π, and θ π. The integrand x +y +z is written ρ in spherical coordinates. We integrate E π x + y + z dv π π π π ρ ρ sin φ dρ dφ dθ ( ) sin φ dφ dθ [ cos φ ] φπ φ dθ π π π π π sin φ dφ dθ dθ (π ) π. sin φ dρ dφ dθ 7

8 Problem 9 Which of the following polar equations defines the circle below? A) r sin(θ) B) r cos(θ) C) r sin(θ) cos(θ) D) r θ cos(θ) E) r θ sin(θ) F) r + sin(θ) G) r sin(θ) H) r θ There are several ways to solve this problem. We will provide you with two of them:. The point (x, y) (, ) is on the circle. In polar coordinates this point can be written (r, θ) (, π) or (r, θ) (, π ). The equation in answer C) is the only one which this point satisfies.. The circle in the picture has center (, ) and radius. So the equation in terms of x and y is (x + ) + (y ). Rewriting to polar coordinates yields (x + ) + (y ) x + x + + y y + x + y y x Convert to polar coordinates r r sin(θ) r cos(θ) r sin(θ) cos(θ). 8

9 Problem Evaluate x cos ( y ) dy dx. by changing the order of integration. ( ) A) cos() B) cos() C) ( ) sin() D) sin() E) cos() F) cos() G) sin() H) sin() To change the order of integration, we first draw the region we are integrating over x Here we see that if we want to integrate with respect to x first, we need to let x run between and y, for y between and. We can then calculate the integral, using the substitution u y, x cos ( y ) dy dx y cos ( y ) dx dy cos u du [ sin u ] (y ) cos ( y ) dy ( sin() ) sin(). 9

10 Problem Find the tangent vector of the curve with parametric equations x(t) t sin ( πθ) dθ, and y(t) t cos ( πθ) dθ at the point corresponding to t. A), B), C), D), E), F), G), H), The Fundamental Theorem of Calculus tells us that and substituting t gives r (t) x (t), y (t) sin( πt ), cos( πt ) r () sin( π), cos( π),. Problem Describe the boundary of the solid region E that we integrate over in f(x, y, z) dv x x y E x x +y f(x, y, z) dz dy dx. A) A paraboloid and a plane B) A sphere C) A cone and a half-sphere D) A paraboloid and a half-sphere E) A cone and a plane F) An ellipsoid G) Two cones H) A cone and a paraboloid The region E is described by E {(x, y, z): x + y z x y, x y x, x }. The boundary of E is described by the surfaces z x + y, which is a cone, and z x y, which is a paraboloid.

11 Problem Let D be the region bounded by the circle x + y x. Evaluate ( x + y ) y da. D A) B) C) D) E) π F) π G) π H) π The expression x + y appears twice in this problem, so it would probably be best to use polar coordinates. First of all, let s rewrite the equation of the circle: x + y x x x + + y (x ) + y. So the circle has center (, ) and radius Let us rewrite the equation in terms of polar coordinates: x + y x r r cos θ r cos θ. There are different possible values for the bounds of r and θ for the region within this circle, but the most convenient is π θ π and r cos θ. We can evaluate the integral: D (x + y )y da π π π π cos θ r r sin θ r dr dθ [ 5 r5 sin θ ] r cos θ r dθ π π π π cos θ 5 cos5 θ sin θ dθ. r sin θ dr dθ

12 If we now do the substitution u cos θ, then du sin θ dθ, whence π π Here s another way to see it: π π 5 cos5 θ sin θ dθ 5 cos5 θ sin θ dθ [ cos6 θ ] π π 5 u5 du [ u6].. Alternatively, we can just argue that 5 cos5 θ sin θ is an odd function of θ, so the integral must be equal to zero. Problem Let S be the surface of revolution obtained by rotating the curve y x, x 8, about the x-axis. Compute the surface area of S. A) 5π B) 5π C) π D) 5π E) π F) 5π 5 5 G) 5π H) 5π To calculate the surface area of a surface of revolution, we can use the formula or the more general area(s) π b a area(s) f(x) + [f (x)] dx, D r x r θ da. In the latter case, we use the vector function r(x, θ) x, f(x) cos θ, f(x) sin θ. Let us use the first formula. We calculate the derivative of f, f (x) x x. Then area(s) π 8 8 x + x dx π x + dx π [ (x + ) x + ] 8 8π ( 9 ) 5π.

13 Problem 5 Which vector field do you see in the picture? y x A) F (x, y) x, y B) F (x, y) x, y C) F (x, y) x, y D) F (x, y) x, y E) F (x, y) y, x F) F (x, y) y, x G) F (x, y) y, x H) F (x, y) y, x On the positive x-axis (where y ), the arrows are horizontal and pointing in the positive direction of x. Using this argument, and looking at the formulas for the vector fields in the eight answers, the only possible correct answers are A) and C). On the positive y-axis (where x ), the arrows are vertical and pointing in the negative y-direction. This leaves C) as the correct answer.

14 9 Hjelle Name: Student-ID: Section: Berson Hjelle The following problem will be hand-graded. To earn full credit you need to justify your answers. Problem 6 a) Calculate 9 x 9 x x + y dy dx. b) Find the volume of the solid bounded by the cone z x + y, the cylinder x + y 9 and the xy-plane. a) We recognize that the integral is over the disk D with center at the origin and radius. Also, the integrand can be expressed in polar coordinates as r. Therefore 9 x 9 x x + y dy dx. π ( 7 9) π ( r) r dr dθ π dθ 9 (π ) 9π. [ r r] dθ b) To find the volume, we need to integrate the height of the cone over the projection of the solid to the xy-plane. Since the solid lies inside the cylinder x + y 9, the projection is the disk with center at the origin and radius. The height is given by z x + y r. Thus, V π r r dr dθ. This is part of the integral we calculated above, so π V 9 dθ 8π. Alternatively, we could have seen that in a), we calculated the volume of the given cylinder cut off at height and height, with everything under the cone removed. Since the volume of the cylinder is V π h r π 7π, the volume under the cone is 7π 9π 8π.