We get a r b first. We can now observe that y 0 implies 0 θ π. Also y = x corresponds to θ = π 4 and y = x corresponds to θ = 3π 4.

Transcription

1 iscussion 24 solution Thursday, April 4th topic: polar coordinates The trick to working with the polar coordinate change of variables is that we don t worry about the substitution r(x, y) and θ(x, y) to determine. We are supposed to recognize from previous work in calculus. For example, the region in xy-space that describes the the first quadrant portion of the disk x 2 + y 2 is the polar rectangle r, θ π/2.. escribe the polar rectangle (the region in rθ-space) that describes the following in xy-space under the polar coordinates change of variable x(r, θ) : r cos θ and y(r, θ) : r sin θ. (a) Let be the disk x 2 + y 2 4. Replacing x r cos θ and y r sin θ, we get r 2 cos 2 θ + r 2 sin 2 θ 4. This is the same as r 2 4. Since there is no restriction on θ and r is always greater than or equal to zero, we conclude r 2, θ 2π. (b) Let be the region bounded by the arcs of the circles x 2 + y 2 a 2, x 2 + y 2 b 2 (where < a < b), the lines y x, y x with y. We obtain the following r 2 a 2, r 2 b 2, r sin θ r cos θ, r sin θ r cos θ, and r sin θ. We get a r b first. We can now observe that y implies θ π. Also y x corresponds to θ π 4 and y x corresponds to θ 3π 4. Therefore a r b, π 4 θ 3π 4. (c) Let be the region in the first quadrant bounded by x 2 + y 2 4, y x/ 3 and y.

2 The first quadrant corresponds to θ π 2. x2 + y 2 4 becomes r 2 4. Finally y x/ 3 corresponds to θ π 6 (because tan π 6 3 ) and y corresponds to θ. Hence r 2, θ π Evaluate x 2 + y 2 + da, where is the portion of the unit disk in the first quadrant. We change to polar coordinate and evaluate. The Jacobian for polar coordinate transform is r. x2 + y 2 + da π/2 r2 + r drdθ π/2 π/2 π 6 ( 8 ) 3 (r2 + ) 3/2 dθ 8 dθ 3 3. Evaluate the integral xy da where is the region in the first quadrant bounded by x 2 + y 2 4, y x/ 3 and y. We change to polar coordinate and evaluate. Note that the region is the same as in (c). xy da π/6 2 π/6 2 π/6 π/6 2 sin 2 θ π/6 /2 (r cos θ)(r sin θ) r drdθ r 3 cos θ sin θ drdθ r 4 4 cos θ sin θ 2 dθ 4 cos θ sin θ dθ

3 topic: general change of variables 4. Let T (u, v) : (u 2 v 2, 2uv). (a) Show that the image of the square {(u, v) u, v } under T is the region in xy-space bound by the x-axis and by the parabolas y 2 4 4x and y x. (Hint: parametrize the four boundary lines of and draw the image of this curves under T.) Part : Consider the curve C (t) (t, ), t. C (t) is the line segment of the bottom of the square. T (C (t)) describes a curve in xy-space which is going to be part of the boundary of the image of. T (C (t)) (t 2 2, 2 t ) (t 2, ). As t increases from to, T (C (t)) goes from (,) to (,) along the curve described by y. Part 2 : Consider the curve C 2 (t) (, t), t. C 2 (t) is the line segment of the leftside of the square. T (C 2 (t)) ( 2 t 2, 2 t) ( t 2, ). As t increases from to, T (C 2 (t)) goes from (,) to (-,) along the curve described by y. Part 3 : Consider the curve C 3 (t) (t, ), t. C 3 (t) is the line segment of the top of the square. T (C 3 (t)) (t 2 2, 2 t ) (t 2, 2t). As t increases from to, T (C 3 (t)) goes from (-,) to (,2) along the curve described by x t 2 (y/2) 2 (which is y x).

4 Part 4 : Consider the curve C 4 (t) (, t), t. C 4 (t) is the line segment of the rightside of the square. T (C 4 (t)) ( 2 t 2, 2 t) ( t 2, 2t). As t increases from to, T (C 4 (t)) goes from (,) to (,2) along the curve described by x t 2 (y/2) 2 (which is y 2 4 4x). We combine our results. We found out that the image of under T is bounded by the three curves y, y 2 4 4x, and y x. Also the corners of the square goes to the points (,), (,), (,2), and (-,). (b) Compute the area of the region in xy-space bound by the x-axis and by the parabolas y 2 4 4x and y x. (Hint: Note T ( ). I wouldn t calculate this directly in xy-space. Use a Jacobian and the Change of Variables Theorem.) We compute the Jacobian first. x x u v 2u 2v y y 2v 2u 4u2 + 4v 2 u v The area is obtained by integrating. da (4u 2 + 4v 2 ) dudv 4u 2 + 4v 2 dudv v2 dv a change of variables dictated by integrand Let e (x+y)/(x y) da where is the trapezoidal region with vertices (, ), (2, ), (, 2) and (, ). (a) Take seconds and convince yourself that you do NOT know a useful antiderivative to f(x, y) : e ( ) x+y x y in either x or y.

5 ( seconds of silence) (b) o you know an antiderivative to f(u, v) : e u/v in either u or v? efine a change of variables u(x, y) and v(x, y) which changes the integrand to e u/v. We know how to integrate e u/v with respect to u. efine u(x, y) x + y and v(x, y) x y. (c) Find the image of under the substitution u(x, y) and v(x, y). (Remark: The image will also be a trapezoidal region. Sadly, not all are rectangles.) is bounded by the lines x y, x y 2, y, and x. Notice that v x y is bounded between and 2. y corresponds to (u v)/2 and x corresponds to (u + v)/2. Therefore the image of under the substitution is the trapezoidal region bounded by the lines v, v 2, v u, and v u. (d) Find the inverse transformation T (u, v) : (x(u, v), y(u, v)) and determine the Jacobian (e) T (u, v) (x, y) (u, v). ( u + v 2, u v ). The Jacobian is Use the change of variables theorem and calculate e x+y x y da. (Hint: Even though is not a rectangle, it is a fairly easy region to describe as a single double integral in the order of dudv. Why would you not want to do dvdu?) We would not want dvdu because in 5(b) we already decided the order we

6 can evaluate. e x+y x y da e u 2 v 2 2 v 2 da v 2 e u v dudv v 2 e u v v v dv v 2 (e e ) dv 4 4 (e e ) 3e 3e 4