Double integrals using polar coordinates Let s look at a motivating example:

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1 Double integrals using polar coordinates Let s look at a motivating example: Example Evaluate e x +y da where is the bounded region in the first quadrant (x 0, y 0) between the circles of radius 1 and (1 apple x + y apple 4). Analysis: Sketch. Figure: egion : x 0, y 0, 1 apple x + y apple 4 Horizontal and vertical slices won t work, since we cannot do Z e x +constant dx or Z e y +constant dy. Slicing radially simplifies the problem. We use polar coordinates (r, ). Math 67 (University of Calgary) Fall 015, Winter / 1

2 Area Element in Polar Coordinates Z Question: How can we write f (x, y) da in terms of r and? We cannot just use da = drd, sincetheequalityissimplynottrue:da 6= drd. It turns out that the correct expression for da is da = rdrd. Let s see why. Figure: Area Element in Polar Coordinates da = rdrd Math 67 (University of Calgary) Fall 015, Winter 016 / 1

3 Area element in polar coordinates Suppose a region is described in polar coordinates by 0 apple g( ) apple r apple h( ), apple apple. Chop up into a mesh of radial lines ( =constants)andarcs(r =constants) Figure: Area Element in Polar Coordinates da = rdrd In the Cartesian coordinate system, A = x y and the area element is da = dxdy. For the small pieces in the mesh of radial lines and arcs, we need to find A. Consider the small piece bounded by radial lines = j 1 and = j and arcs r = r i 1 and r = r i. By the arc length formula L = r, theouterandinnercirculararcsofthepiecehavelengthsr i j and r i 1 j,where j = j j 1. As we chop into a finer mesh, the lengths of the outer and inner circular arcs are approximately the same, and the area of the piece is approximately the area of a rectangle with base r i j and height r i = r i r i 1. So, A ij = r i r i j. The area element in polar coordinates is, therefore, da = rdrd. Math 67 (University of Calgary) Fall 015, Winter / 1

4 Double integrals and polar coordinates Let f be a continuous function on a bounded region (in the xy-plane) that admits a polar coordinate description = {(x, y) 0 apple g( ) apple r apple h( ), apple apple }, where 0 < apple. Then f (x, y) da = Z = = Z r=h( ) r=g( ) f (r cos, r sin ) rdrd. Proof (Outline): We follow the same limiting process in the definition X X f (x, y) da = lim max x i, y j!0 i j f (x ij, y ij ) A ij. The sample point (x ij, y ij )inpolarcoordinatesis(r ij cos ij, r ij sin ij ). The area of the small pieces in the mesh is A ij = r i r i j. To describe the diminishing dimensions of the small pieces, the polar coordinates version of max x i, y j! 0ismax r i, j! 0. Putting all these together in a limiting process: mx f (x, y) da = lim max r i, j!0 i=1 nx j=1 f (r j cos ij, r ij sin ij ) r i r i j = Z = = Z r=h( ) r=g( ) f (r cos, r sin ) rdrd Math 67 (University of Calgary) Fall 015, Winter / 1

5 Setting up iterated integrals in polar coordinates To set up the integral, we find the polar description of the region: Holding fixed, draw a radial line from the origin through at an angle of. The radial line crosses on its boundary. The inner function is r = g( ) wheretheradialsliceenters, and the outer function is r = h( ) wheretheradialsliceleaves. Imagine rotating the radial line about the origin sweeping out the region. Theleastangleis, andthe greatest angle to cover is. Let s look at our original example: Example Evaluate e x +y da where is the bounded region in the first quadrant (x 0, y 0) between the circles of radius 1 and (1 apple x + y apple 4). Figure: egion : x 0, y 0, 1 apple x + y apple 4 Math 67 (University of Calgary) Fall 015, Winter / 1

6 Solution: Sketch. Figure: egion :1apple r apple, 0 apple apple / Slice radially at 0 apple apple. The inner boundary curve is r =1andtheouterboundarycurveisr =. Using x + y = r, e x +y da = (using u = r, du =rdr) = = Z = / Z r= =0 Z = / =0 e 4 r=1 e r! e e r rdrd r= r=1 / 0 d = Z / 0 = e4 e e 4! e 1! d Math 67 (University of Calgary) Fall 015, Winter / 1

7 Example Find xe y da, where is the region in the first quadrant bounded inside the circle x + y =9. Solution: Using the polar coordinate system, the region is 0 apple r apple 3, 0 apple apple. It turns out to be easier to integrate with respect to first: Z r=3 xe y da = Z = / =0 (r cos )e r sin rd dr (use u = r sin, du = r cos d ) = Z r=3 re r sin = / dr =0 = Z r=3 re r r dr (by parts) = re r e r r = 3e 3 e 3 9 = e 3 7.! r=3 (0 1 0) Math 67 (University of Calgary) Fall 015, Winter / 1

8 Example Find (x + y) da, where is the region enclosed inside the circle (x 1) + y =1. Solution: ecall that the polar equation for this circle is r =cos, apple apple. Figure: Curve on the right bounds region :0apple r apple a cos, apple apple xda = Z = / = / Z r= cos Z = / (r cos ) rdr d = = / (cos ) r3 3! r= cos d = 8 3 Z / / cos 4 d =... =. Math 67 (University of Calgary) Fall 015, Winter / 1

9 Further examples/exercises Evaluate the following double integrals: 1 (x + y) da, whered is the region in the first quadrant inside the circle x + y = D D p x + y da, whered is the region bounded between the circles x + y = a and x + y = b (0 < a < b). y tan 1 da, whered is the region (x, y) :9apple x + y apple 16, y applex. x D D x yda,whered is the region bounded inside the curve r =sin. yda,where is the overlap between the disks x + y apple 1andx +(y 1) apple 1. xda,where is the overlap between the disks x + y apple 1andx +(y 1) apple 1. 7 Find the volume of the solid bounded by the paraboloid z =4 x y and the xy-plane. 8 Find the volume of the solid bounded by the cylinder (x 1) + y =1,theplanez =0andthecone z = p x + y. Math 67 (University of Calgary) Fall 015, Winter / 1

10 Extra Examples ecall: The area of a bounded region D is equal to D 1 da. Proof: The double integral computes the net volume under the constant function f (x, y) =1overD, whichis equal to the area of D times the height 1. Example Find the area of = (x, y) 1 apple x + y apple 16 Solution: Sketch. Figure: egion :1apple r apple 4, 0 apple apple Slice radially at (0 apple apple ). Each slice enters at r =1andleavesatr =4. 1 da = Z = Z r=4 =0 r=1 1rdrd = Z = =0 r r=4 r=1 d = Z 0 15 d =15 Math 67 (University of Calgary) Fall 015, Winter / 1

11 ASpecialCase If r ranges from 0 to f ( ) and from to,thentheareaoftheregionis: A = Z = = Z r=f( ) 1rdrd = Z = = r r=f( ) d = 1 Z = = (f ( )) d We have seen this formula previously, but now you know more about where it comes from. We have also done an example very similar to the following. Can you solve it without looking at the solution? Example Find the area of one petal of the rose r =sin(3 ). Solution: Sketch the rose. [Try to do it yourself! Check your sketch by using Wolfram Alpha.] Slice radially at. Eachslicerangesfromr =0tor =sin(3 ). As increases we travel around the petal. We want the value of where we end up back at the origin (i.e., the range of for the first loop). For what is r =0? r =sin(3 ) =0 () 3 = n () = n 3, n =0, ±1, ±,... For from 0 to 3 we travel around the first loop. Math 67 (University of Calgary) Fall 015, Winter / 1

12 Using the area formula for the special case: A = 1 Z = /3 =0 (sin(3 )) d = 1 Z /3 0 1 cos(6 ) d = 1 4 = 1 4 Z /3 0 (1 cos(6 )) d sin(6 ) 6 /3 0 = 1 Math 67 (University of Calgary) Fall 015, Winter / 1