Math 261 Solutions To Sample Exam 2 Problems

Transcription

1 Solutions to Sample Eam Problems Math 6 Math 6 Solutions To Sample Eam Problems. Given to the right is the graph of a portion of four curves:,, and + 4. Note that these curves divide the plane into separate regions, which have been marked on the diagram. (a) Write R da as an iterated integral, both in the dd order and in the d d order. Then, evaluate one of the two integrals. (,) (,) (,) R R R (b) Set up R da as an iterated integral, both in the dd order and in the d d order. (c) Use polar coordinates to evaluate R ( + )da. (d) Use polar coordinates to evaluate R da, where R is the region formed b combining the regions R and R. (e) Set up, but do not evaluate, an iterated integral equivalent to da, where S is the region formed S b combining the regions R and R. Use whatever coordinate sstem ou think is easiest. (a) Integrating in the d d direction means that we will be integrating in the - direction first, meaning that we need to find the curves at the bottom and top of the region R. From the diagram, we see that the curve on the bottom is the line, and the curve on the top is the circle + 4. Solving for, we have + 4 4, so R da 4 d d. Integrating in the dd direction means that we will be integrating in the -direction first, meaning that we need to find the curves on the left and right of the region R. From the diagram above, we see that the curve on the left is the line, and the curve on the right is the circle + 4. Solving for, we have + 4 4, so Calculating, we see that R da 4 dd. (, ), 4, 4 d d 4 d (4 )d ) (4 5. (b) As above, integrating in the d d direction means that we need to locate the top and bottom curves and solve for. The top curve is the line, while for the bottom curve we have. Therefore, R da (/ ) d d.,,, To treat the dd direction, note that the curve on the left side of R is, while for the curve on the right side we have. Therefore, R da d d. (c) Since the point (, ) at the upper right-hand corner of R is given b (, /6) in polar coordinates, it follows that R {(r, θ) : r, / θ /6}. Therefore, we have ( + 6 )da r r dθ dr (since + r and da r dθ dr) R

2 Solutions to Sample Eam Problems Math 6 ( 6 + ) r dr 8. (d) We have R {(r, θ) : r, /6 θ /}, so R () da 6 6 r cosθ r dr dθ (4r ) cosθ dθ sinθ / /6 6. (since r cosθ and da r dr dθ) (e) Since the region S is bounded b onl one curve on the left ( ) and onl one curve on the right ( 4 ), the integration will be the simplest in the dd order. We therefore have S da 4 d d.,, 4 Other equivalent answers require two integrals. Here are two alternate answers that are correct, though the are more complicated: 6 4 d d + r cosθ r dr dθ d d, sin θ r cosθ r dr dθ. or. Pictured to the right is a thin metal plate in the shape of a quarter circle placed in the first quadrant. All distances are measured in inches. (a) Find the mass of the plate if its densit is a uniform grams per square inch. (b) Find the mass of the plate if its densit is given b σ(, ) + grams per square inch. Do NOT use our calculator to evaluate the integral. First, we recall that, if σ(, ) represents our densit function, then Mass σ(, )da, which makes sense because σ has units of grams/in and da has units of in, indicating that the above integral will have units of grams. E (a) We could calculate an integral to find the mass, but since the densit is constant for part (a), we simpl have Mass (Densit)(Area of Plate) ()(/4) 5 grams. (b) In this case, we need to integrate since the densit is not constant. Therefore, the mass is given b E + da E + da 5 grams. r r dr dθ

3 Solutions to Sample Eam Problems Math 6. Evaluate dv, where W is the first-octant portion of the solid sphere of radius centered at the origin. W Do this in was: using clindrical coordinates and using spherical coordinates. (a) In spherical coordinates, we have ρ cosφ, and the equation of our sphere becomes ρ. Therefore, we have dv (ρ cosφ)p sinφdθ dφ dρ ρ sinφcos φdφ dρ W (b) In clindrical coordinates, we have 4. ρ sin φ ρ dρ dρ r + 4 ± 4 r, and since for the portion of the sphere we are interested in, the equation for the surface of the sphere becomes 4 r. Therefore, we have W dv 4 r r dθ d dr 4 r r d dr 4. r 4 r r(4 r )dr dr 4. Consider the solid region in the st octant that lies below the plane + 4 and inside the clinder + 4 (see diagram to the right). Set up, but do not evaluate, iterated integrals that give the volume of this region in two was: in the d d d order and in the dd d order. First, we tackle the d d d order of integration. The surface on the left hand side of the region is the plane, and the surface on the right hand side of the region is the plane + 4, which is equivalent to (4 ). To the right, note that the -projection of this solid is shown, where the - and -intercepts have been found b setting and in the equation + 4 and observing that and are both positive. Therefore, we have 4 Volume 4 (4 ) d d d.

4 Solutions to Sample Eam Problems Math 6 4 Net, we tackle the dd d order of integration. In this case, the surface on the back side of the region is the plane, and the surface on the front side of the region is the clinder + 4, which is equivalent to 4 since we are in the first octant. To the right, note that the -projection of this solid is shown, where the various - and -intercepts have been found b equating appropriate surfaces. Because there are two different top curves in the projection picture, we will need two separate integrals to describe the integral over the projected region shown above. Therefore, we obtain 4 Volume 4 dd d dd d. 5. Find parameteriations for the following curves. (a) The circle of radius 4, parallel to the -plane and centered at (,, ). (b) The line segment starting at (,, 4) and ending at (, 5, ). (a) 4 cost,, 4 sint, for t. (b) First, we note that the vector v ( ) i + (5 ) j + ( 4) k i + 4 j 5 k is parallel to the line segment. Therefore, a parameteriation for this segment is given b 6. The National Park Service has conducted a cactus census in Ariona. In order to collect data, the service divided the desert into a grid whose lines are miles apart in each direction. The entries in the table to the right represent thousands of cacti per square mile at grid points in Saguaro East National Park, where the visitor center is located at (, ). + t, + 4t, 4 5t, where t. Distance, (miles east) Distance, (miles south) (a) Let f(, ) be the number of cacti (in thousands) per square mile miles east and miles south of the visitor center. Eplain what f(, )d d represents in the contet of this problem. (b) Give the best estimate ou can based on the data of the number of cacti in the b test area. (a) First, we eamine the units of the various pieces of our integral. The function f(, ) is measured in thousands of cacti per square mile, and d and d are both measured in miles. Therefore, the units of the integral are given b thousands of cacti mi mi mi thousands of cacti. Therefore, this integral represents the population of cacti (in thousands) in the b region of the desert described b the provided table. (b) To estimate the number of cacti, we need to estimate the population densit in each subsquare represented b the table and multipl b the area. For eample, in the subsquare described b and (the upper left square described b the table), there are four different population densities we could use: 8.5, 8., 9.5, or.6. These are just the numbers that are given at the four corners of the subsquare. Therefore, a reasonable estimate of the population densit for this subsquare would be thousand cacti per square mile

5 Solutions to Sample Eam Problems Math 6 5 Doing this for each of the 9 subsquares described from the table, we obtain the population densities described b the diagram above. To get a population, we need to take each of these population densities and multipl b the area of the corresponding subsquare, which is square miles for each of them. Therefore, our best estimate of the cactus population is given b ( ) 867 thousand cacti, or about 8.67 million cacti. 7. Given below are three vector fields, F, F, and F. Let C and C denote circles of radius and, respectivel, both centered at (, ) and oriented counterclockwise. Also, let C denote the line segment starting at (, ) and ending at (, ), and let C 4 denote the line segment starting at (, ) and ending at (, ). F (, ) F (, ) F (, ) (a) For each of the four curves C, C, C, and C 4, and for each of the vector fields F, F, and F, decide whether C F d r appears to be positive, negative, or ero. Note that there are different integrals to be eamined. (b) Which is larger, C F d r or C F d r? Eplain. (a) Vector Field F : F d r >, F d r >, F d r, F d r <. C C C C 4 Vector Field F : F d r <, F d r <, F d r >, F d r >. C C C C 4 Vector Field F : F d r, F d r, F d r <, F d r. C C C C 4 (b) All the vectors in the vector field appear to have the same length. Therefore, since both curves go directl with the flow of the vector field but C is a longer path than C, we conclude that C F d r < C F d r. 8. Let F i + ( + ) j, and let G ( + ) i + ( ) j, where C is the curve consisting of the circle of radius, centered at the origin and oriented counterclockwise, and where C is the curve consisting of the line segment from (, ) to (, ) followed b the line segment from (, ) to (, ). (a) Calculate the line integral of F over C. (b) Calculate the line integral of G over C. (c) Calculate the line integral of F over C. (d) Calculate the line integral of G over C. Before we begin, let us check to determine whether F or G are conservative vector fields. Since ( ) 4 ( + ), we see that F is a conservative vector field. To find a potential function f

6 Solutions to Sample Eam Problems Math 6 6 for F, we integrate as follows: f(, ) d f(, ) ( + )d + P() + + Q() Therefore, f(, ) + is a potential function for F. On the other hand, since ( + ) is not equal to ( ), we see that G is not a conservative vector field. (a) Since C is a closed curve and F is a conservative vector field, we know that C F d r without doing an calculations. So our final answer is. (b) Since G is not a conservative vector field, we must do this integral b parameteriing C. We can represent C b the parametric curve r(t) cost i + sint j, where t. Therefore, we have C G d r so our final answer is 8. G( cost, sin t) ( sint i + cost j)dt ( sin t + cost) i + ( sin t cost) j) ( sint i + cost j)dt [ 4(sin t + cos t) 4 sintcost + 4 sintcost] dt ( 4) dt, (c) Since F is conservative, we can use the potential function f(, ) + that we calculated above, and the Fundamental Theorem of Calculus for Line Integrals as follows: C F d r f(, ) f(, ) Therefore, our final answer is. (d) Since G is not conservative, we must parameterie the two line segments comprising C and calculate this line integral using brute force. The line segment from (, ) to (, ) can be parameteried as r (t) t i + t j, where t, and the line segment from (, ) to (, ) can be parameteried as r (t) ( + t) i + j, where t. Therefore, we have so our final answer is 7. C G d r G( r (t)) r (t)dt + G( r (t)) r (t)dt (t i + j) ( i + j)dt + (t + ( + t))dt (6t + 4)dt 7, (( + t) i + t j) idt 9. Let F i+ sin() j+ sin() k. Calculate C F d r, where C is the path from A (,, ) to B (,, ) shown in the figure to the right. B A

7 Solutions to Sample Eam Problems Math 6 7 Calculating, we observe that curlf, so the vector field F is conservative and we can use the Fundamental Theorem of Calculus for Line Integrals, provided that we can find a potential function, which we do below: f(,, ) d + p(, ) f(,, ) sin()d cos() + q(, ) f(,, ) sin()d cos() + r(, ) Therefore, f(,, ) cos() is a potential function for F, so b the Fundamental Theorem of Calculus for Line Integrals, we have F d r f(,, ) f(,, ) (9 cos) ( cos) cos. C