Dr. Z’s Math251 Handout #15.4 [Double Integrals in Polar Coordinates]By Doron ZeilbergerProblem Type 15.4a: Evaluate the integralZ ZDF (x, y) dA ,where D is a region best described in polar coordinates,D = {(r, θ) |α ≤ θ ≤ β , h1(θ) ≤ r ≤ h2(θ) } .Example Problem 15.4a: Evaluate the integralZ ZDe−x2−y2dA ,where D is the region bounded by the semi-circle x =p25 −y2and the y-axis.Steps Example1. Draw the region and express it, if pos-sible and convenient, asD ={(r, θ) |α ≤ θ ≤ β , h1(θ) ≤ r ≤ h2(θ) } .Of course, in many problems, the h1(θ)and/or h2(θ) may be plain numbers (i.e.not involve θ).1. This is a semi-circle, i.e. half a cir-cle, center origin, radius 5, and since it isbounded by the y-axis, and x ≥ 0, it isthe right half[Had it been x = −p25 −y2it wouldhave been the left-half. Had it been y =√25 −x2it would have been the upper-half. Had it been y = −√25 −x2it wouldhave been the lower-half.]Since it is the right-half, θ ranges fromθ = −π/2 (the downwards direction) toθ = π/2 (the upwards direction). Foreach ray θ = θ0, r, the distance from theorigin, ranges from r = 0 to r = 5 (andindeed does not depend on θ in this prob-lem). So our region phrased in polar co-ordinates is:D = {(r, θ) |−π/2 ≤ θ ≤ π/2 , 0 ≤ r ≤ 5 } .12. Rewrite the area integralZ ZDF (x, y) dA ,in polar coordinates by replacingx by r cos θ, y by r sin θ, dA by r dr dθ.[shortcut: Whenever you see x2+y2youcan replace it by r2.]Write it as an iterated integralZ ZDF (x, y) dA =ZβαZh2(θ)h1(θ)F (r cos θ , r sin θ) r dr dθ ,with the θ-integral being at the outsideand the r-integral being in the inside.2.Z ZDe−x2−y2dA=Zπ/2−π/2Z50e−r2rdr dθ .3. Evaluate this iterated integral by firstdoing the inner-integral (possibly gettingan expression in θ, or just a number), andthen the outer integral.3. The inside integral is (do the change-of-variable u = −r2):Z50e−r2rdr = (−1/2)e−r250= (1−e−25)/2 ,and the whole double-integral isZπ/2−π/2Z50e−r2rdr dθ=Zπ/2−π/2Z50e−r2rdrdθ=Zπ/2−π/2[(1−e−25)/2] dθ = (1−e−25)/2Zπ/2−π/2dθ =[(1−e−25)/2][π/2−(−π/2)] = π(1−e−25)/2 .Ans.: π(1 − e−25)/2 .2Problem Type 15.4b: Find the volume of the solid above the surface z = f(x, y) and below thesurface z = g(x, y) .Example Problem 15.4b: Find the volume of the solid above the cone z =px2+ y2and belowthe sphere x2+ y2+ z2= 2 .Steps Example1. Find the “floor”, let’s call it D, bysetting f (x, y) = g(x, y) (or if convenientalready convert to polar coordinates).1. In polar coordinates, the two surfacesare z = r and z =√2 −r2. Settingthem equal gives r =√2 −r2. Squar-ing both sides gives r2= 2 − r2, whichgives 2r2= 2, which gives r2= 1 andso r = ±1. But r is never negative, sor = −1 is nonsense. Hence the “floor”, D,is the region bounded by the circle r = 1,or, if you wish, the disk r ≤ 1.SoD = {(r, θ)|0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π} .2. The volume is the area integral ofTOP-BOTTOMZ ZD[f(x, y) − g(x, y)] dA .Set it up. Then convert it to polar-coordinates.2. The bottom is z =px2+ y2, and inpolar z = r, and the top is x2+ y2+z2= 2 which is z =p2 −x2− y2andin polar z =√2 −r2. So the volume inpolar coordinates isZ2π0Z10[p2 −r2− r] r dr dθ .33. Evaluate the iterated integral. Firstdo the inner integral (w.r.t. to r) gettingan expression in θ (or just a number), andthen do the outer integral.3. The inner integral isZ10[p2 −r2−r] r dr =Z10[rp2 −r2−r2] dr=Z10r(2 − r2)1/2dr −Z10r2dr= −(1/3)(2 −r2)3/210−r3/310= −(1/3)(2 −r2)3/210−r3/310= −(1/3)[(2 −12)3/2− (2 −02)3/2] −1/3= [23/2− 2]/3 = (2√2 −1)/3 .The whole integral is thus:Z2π0Z10[p2 −r2− r] r dr dθ=Z2π0Z10[p2 −r2− r] r drdθ=Z2π0(2√2 −1)/3 dθ= 2π(2√2 −1)/3 .Ans.: The volume is 2π(2√2 −1)/3.Problem Type 15.4c: Evaluate the iterated integral by converting to polar coordinates.ZbaZf2(y)f1(y)F (x, y) dx dyExample Problem 15.4c: Evaluate the iterated integral by converting to polar coordinates.Z30Z√9−y2−√9−y2x2y dx dySteps Example41. By looking at the limits of integrationof the outer and inner integral signs, fig-ure out the region D.D = {(x, y) |a ≤ y ≤ b , f1(y) ≤ x ≤ f2(y) } .Draw this region, and express it in polarcoordinatesD = {(r, θ) |α ≤ θ ≤ β , g1(θ) ≤ r ≤ g2(θ) }1. Our region is:D = {(x, y) |0 ≤ y ≤ 3 , −p9 −y2≤ x ≤p9 −y2} .Drawing it (do it!), we see that this is theupper-half of the circle whose center is theorigin and whose radius is 3. In polar co-ordiantes it is:D = {(r, θ) |0 ≤ θ ≤ π , 0 ≤ r ≤ 3 } .2. Write the iterated integral as an areaintegral, then convert it to an iterated in-tegral in polar coordinates. Use the “disc-tionary” x = r cos θ y = r sin θ dx dy =r dr dθ.2.Z30Z√9−y2−√9−y2x2y dx dy=Zπ0Z30(r cos θ)2(r sin θ)r dr dθ=Zπ0Z30r4sin θ cos2θ dr dθ .53. Evaluate that iterated integral by do-ing the inner integral first, and then theouter integral.3. The inner integral isZ30r4sin θ cos2θ dr = sin θ cos2θZ30r4dr= sin θ cos2θr5530=2435sin θ cos2θ .The outer integral is:Zπ0Z30r4sin θ cos2θ dr dθ=Zπ0Z30r4sin θ cos2θ drdθ=Zπ02435cos2θ sin θ dθ =2435Zπ0cos2θ sin θ dθ=2435·−cos3θ3π0=815· (−cos3(π)−−cos3(0))