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.
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 6. 2. 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
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 2 4 + 4x. (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 2 4 + 4x).
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 2 4 + 4x. 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 2 4 + 4x. (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 4 3 + 4v2 dv 4 3 + 4 3 8 3 5. 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.
( 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 2 2 2 2 2 2. 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
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