Mathematics 5 Solutions for HWK Problem 1. 6. p7. Let D be the unit disk: x + y 1. Evaluate the integral e x +y dxdy by making a change of variables to polar coordinates. D Solution. Step 1. The integrand, e x +y will be replaced by e (r). Step. The expression dx dy or da must be replaced by its polar equivalent rdrdθ. Step. For θ fixed, an r-arrow enters at r and leaves at r 1. Then θ must vary from to π. So the limits of integration will be r 1 and θ π. Step 4. Evaluate the integral using the new variables: e x +y dxdy 1 D e r r dr dθ [ e 1 ] dθ π(e 1) Page 1 of 8 A. Sontag April 5,
Math 5 HWK Solns continued Problem 9. 6. p7. x + y 4. Solution. Evaluate the integral D (x + y ) dxdy, where D is the disk Both the integrand and the region of integration suggest using polar coordinates. Step 1. The integrand, (x + y ), will be replaced by (r ), so by r. Step. The expression dx dy must be replaced by its polar equivalent, namely rdrdθ. Step. The limits of integration will be r and θ π. Step 4. Evaluate the integral using the new variables: (x + y ) dxdy D r r dr dθ r 4 dr dθ 5 dθ 64π 5 Page of 8 A. Sontag April 5,
Math 5 HWK Solns continued Problem 1. 6. p7. Integrate ze x +y over the cylinder x + y 4, z. (Call the cylinder W.) Solution. Both the integrand and the region of integration are well suited to cylindrical coordinates, so make that change of variables. Step 1. The integrand, ze x +y, becomes ze (r). Step. For the expression dv, use its cylindrical equivalent, namely rdrdθdz. Because the solid in question has such a nice cylindrical-coordinate description, we can take the variables in any order. Step. Determine the limits of integration that are needed to describe the cylinder in cylindrical coordinates. In this case whatever order we choose for the three variables, the limits of integration will all be constant: z, r, θ π. Step 4. Evaluate the integral using the new variables (in any order): W ze x +y dv ze r r drdθdz z (e4 1) dθdz πz(e 4 1) dz π(e 4 1) 1 ( ) 5π(e4 1) Page of 8 A. Sontag April 5,
Math 5 HWK Solns continued Problem 19. 6. p7. Integrate x + y + z over the cylinder x + y, z. Solution. Although the integrand is reasonably nice in either cylindrical or spherical coordinates, the region of integration, W say, is more easily described using cylindrical coordinates rather than spherical, so convert to cylindrical coordinates (which are definitely easier to work with here than rectangular). Step 1. In cylindrical coordinates, the integrand x + y + z should be replaced by r + z. Step. For dv use r dr dθ dz. Any order will do. (The region will give constant limits of integration, no matter which order we use, and the new integrand will be easy to integrate, no matter what order we use.) Step. The limits of integration will be r, θ π, z. Step 4. Evaluate the integral using the new coordinates: W x + y + z dv ( r z (z + r )r drdθdz [z + 1]dθdz π(z + 1)dz [ z π + z ] π( 7 + + 8 + ) 1π )] r + r4 dθdz 4 r Page 4 of 8 A. Sontag April 5,
Math 5 HWK Solns continued Problem 1. 6. p7. Using spherical coordinates, evaluate the triple integral where B is the ball x + y + z 1. B dxdydz + x + y + z Solution. Step 1. In spherical coordinates, the integrand 1 + x + y + z is simply 1. + ρ Step. For dv, given as dxdydz, we use the spherical equivalent dv ρ sin φdρdθdφ. Since the region in question has a very nice spherical description, it won t matter what order we use for the variables. Page 5 of 8 A. Sontag April 5,
Math 5 HWK Solns continued Step. In spherical coordinates, the ball B can be described as the set of points whose spherical coordinates satisfy ρ 1, φ π, θ π, and we can integrate in any order using these limits. Step 4. Evaluate the integral using the new variables: I W dxdydz π + x + y + z 1 ρ sin φ + ρ dρdθdφ For the innermost integral, we ll need to use integral tables. For some unknown reason, the formula we need isn t given in the integral tables in the text, but it is in Short Table of Integrals. Formula 15 is u du u ± a u u ± a ±a u ln + u ± a + C which gives us ρ + ρ dρ ρ ρ + ln(ρ + ρ + ) + C so 1 ρ + ρ dρ 1 ln(1 + ) + ln( ) Page 6 of 8 A. Sontag April 5,
Math 5 HWK Solns continued Incorporating this result into what we already had gives I π π 4π π [ ln(1 + ) + ln ] dθdφ [ ln(1 + ) + ln ] dφ sin φ sin φ ( ln(1 + ) + ln ) Problem. 6. p7. Evaluate the triple integral I S dxdydz (x + y + z ) where S is the solid region bounded by (between) the spheres x +y +z a and x +y +z b. (a > b > ) Solution. Both the integrand and the region of integration suggest using spherical coordinates. 1 1 Step 1. The integrand becomes or simply 1 (x + y + z ) (ρ ) ρ. Step. For dxdydz we use the spherical equivalent ρ sin φdρdθdφ. Once again you ll see that it won t matter what order we choose for the variables. Page 7 of 8 A. Sontag April 5,
Math 5 HWK Solns continued Step. This time we want the region between two spheres. The limits of integration will be the same as in Problem 1 for the variables θ and φ, but now ρ will range from the smaller value of b to the larger value a. In other words, we ll use θ π, φ π and b ρ a. Step 4. Evaluate the integral using the new coordinates. I S π a π π π b a 4π ln a b dxdydz (x + y + z ) b 1 ρ ρ sin φ dρdθdφ sin φ ρ dρdθdφ sin φ ln a b dθdφ π sin φ ln a b dφ Page 8 of 8 A. Sontag April 5,