Name (please print) π cos(θ) + sin(θ)dθ

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1 Mathematics Final Eamination Form A December 2, 27 Instructions: Give brief, clear answers. I. Evaluate by changing to polar coordinates: 2 + y 2 2 and above the -ais. + y d 2(2 2 )/3. π 2 (r cos(θ) + r sin(θ))r dr dθ + y d where is the region between 2 + y 2 and π cos(θ) + sin(θ)dθ 2 r 2 dr II. For the function f(,y) ln( 2 +y 2 ), find the maimum rate of change at the point (,2), and the direction in which it occurs. Find the directional derivative of f at (,2) in the direction toward (2,4). We have f(,y) 2 ı + 2y j, so f(,2) 2 2 +y 2 2 +y 2 ı + 4 j. The maimum rate of change at (,2) is f(,2) 2 ı + 4 j 2, and it is in the direction of f(,2). A vector in the direction from (,2) toward (2,4) is ı + 2 j, so a unit vector in that direction is u ı + 2 j. The rate of change in that direction is f(,2) u 2. (Alternatively, one can observe that the given direction is eactly the direction of f(,2), so the rate of change in that direction, as already found, is f(,2) 2.) III. IV. () V. Let be the portion of the sphere of radius a that lies in the first octant. Use the standard parameterization of to calculate (y ı j + k ) d. If is the region θ π/2, φ π/2 in the θφ-plane, then (y ı j + k) d (y ı j + k) ( r φ r θ )d (y ı j + k) a sin(φ)( ı + y j + z k)d asin(φ)z d π/2 dθ π/2 asin(φ)acos(φ)dφ (π/2)a 2 sin 2 (φ)/2 π/2 πa 2 /4 Use the Divergence Theorem to calculate the surface integral ( 2 z 3 ı + 2yz 3 j + z 4 k ) d, where is the surface of the bo with 3, y 2, z. ( 2 z 3 ı + 2yz 3 j + z 4 k ) d 2z 3 + 2z 3 + 4z 3 dv 2 3 8z 3 ddy dz 3 2d 2 E dy 4z 3 dz The radius of a right circular cone is increasing at a rate of 6 in/s while its height is decreasing at a rate of 3 in/s. At what rate is the volume V πr 2 h/3 changing when the radius is and the height is? Using the hain ule, dv dt (πr2 h/3) dr r dt + (πr2 h/3) dr h dt 2πrh dr 3 dt + πr2 dh. Evaluating when 3 dt r and h, we find dv dt π 6 + π ( 3) π. It is increasing at a rate of π 3 3 in 3 /s.

2 Page 2 VI. VII.. 2. Let be the upper half of the sphere of radius 2, that is, the points (,y,z) with 2 + y 2 + z 2 4 and z, and suppose that is oriented with the upward normal. Use tokes Theorem to evaluate curl( 2 e yz ı + y 2 e z j + z 2 e y k) d. The boundary of is the circle of radius 2 in the y-plane, with the positive orientation. By tokes Theorem, curl( 2 e yz ı + y 2 e z j + z 2 e y k) d ( 2 e yz ı + y 2 e z j + z 2 e y k) d r. On, z, so this line integral is (2 ı + y 2 j) d r. Using Green s Theorem, ( 2 ı + y 2 j) d r y 2 2 dd. y D Let be the upper half of the sphere of radius, that is, the points (,y,z) with 2 + y 2 + z 2 and z. Using the geometric interpretation of the surface integral of a vector field as the flu (that is, not by calculation using a parameterization or a formula from the formulas list), eplain each of the following equalities: j d onsider the flow of unit speed in the j direction. j d is the net flow (per unit time) across. ince this flow is horizontal, the amount that flows into the hemisphere on the left side eactly equals the amount that flows out of it on the right, so the net flow is. k d π onsider the flow of unit speed in the k direction. k d 2π is the net flow per unit time across. ince the flow is directly upward, the amount that flows across is the same as the amount that flows across the the unit disk in the y-plane. The amount of vertical flow per unit time is just the area of this disk, which is π. VIII. Verify that the function u cos( at) + ln( + at) is a solution to the wave equation u tt a 2 u. ( (cos( at) + ln( + at)) tt asin( at) + (cos( at) + ln( + at)) a 2 u a 2 ( cos( at) a ) + at ( sin( at) + ) ( + at) 2 a 2 cos( at) a 2 t a2 cos( at) ( + at) 2 and ) + at cos( at) ( + at) 2, so a 2 ( + at) 2 u tt.

3 Page 3 IX. () Let be the portion of the cylinder 2 + z 2 4 that lies between the vertical planes y and y 2. The surface is parameterized by 2cos(θ), y h, z 2sin(θ) for θ 2π and h 2 2cos(θ).. alculate r θ, r h, r h r θ, and r h r θ. r θ 2sin(θ) ı + 2cos(θ) k and r h j. ı j k r h r θ 2cos(θ) ı + 2sin(θ) k, so r h r θ 4cos 2 (θ) + 4sin 2 (θ) 2. 2 sin(θ) 2 cos(θ) 2. alculate d. d r h r θ d 2d, where is the region θ 2π and h 2 cos(θ) in the parameter domain. o we have 3. alculate d k d. cos(θ)2d 2 2cos(θ) 2cos(θ)dhdθ 4cos(θ) 2( + cos(2θ))dθ 4π + 4π. 2cos(θ)(2 2cos(θ))dθ k d k ( r h r θ )d 2 cos(θ) 2sin(θ)cos(θ)dhdθ k (2cos(θ) ı + 2sin(θ) k)d 2sin(θ)cos(θ)(2 cos(θ))dθ 4sin(θ)cos(θ) 2sin(θ)cos 2 (θ)dθ 2sin 2 (θ) 2cos 3 (θ)/3 2sin(θ)cos(θ)d 2π. X. The curl of the vector field y ı z j + k is ı j k. Let be the triangle which is the part of the plane 2 + y + z 2 that lies in the first octant. Give the upward normal, and give its boundary the corresponding positive orientation. Use tokes Theorem to evaluate the line integral (y ı z j+ k) d r. (Hint: the surface integral on is easy to calculate if one uses the definition G d G n d.) A normal vector to is 2 ı+ j+ k. This is upward, since the k-component is positive, and 2 ı+ j+ k 6, so a unit upward normal is n (/ 6)(2 ı + j + k). Using tokes Theorem, we have (y ı + z j + k) d r curl(y ı + z j + k) d ( ı j k) nd ( ı j k) ((/ 6)(2 ı + j + k))d (/ 6)(2 )d.

4 Page 4 XI. In an y-coordinate system, sketch the gradient of the function whose graph is shown to the right. z y y XII. (3) XIII. () A function of the variables, 2, and 3 is given implicitly by Use implicit differentiation to find 3. Applying, we have , so Use the Divergence Theorem to show that if E is a solid with boundary the surface, then ( 3 ı + y 3 j + z 3 k) d always equals the volume of E. ( 3 ı + y 3 j + z 3 k) d E dv dv vol(e). E

5 Page XIV. On two different coordinate systems, graph the following vector fields:. F(,y) ı + y j 2. F(,y) y 2 + y 2 ı y 2 j XV. ketch the region and change the order of integration for e e f(,y)dy d. e ye ln(y) e ln(y) f(,y)ddy.

Name (please print) π cos(θ) + sin(θ)dθ

Name (please print) π cos(θ) + sin(θ)dθ Mathematics 2443-3 Final Eamination Form B December 2, 27 Instructions: Give brief, clear answers. I. Evaluate by changing to polar coordinates: 2 + y 2 3 and above the -ais. + y d 23 3 )/3. π 3 Name please

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