Finl xm olutions, MA 3474 lculus 3 Honors, Fll 28. Find the re of the prt of the sddle surfce z xy/ tht lies inside the cylinder x 2 + y 2 2 in the first positive) octnt; is positive constnt. olution: The surfce is the grph z gx, y) xy/ where x, y) nd is the prt of the disk x 2 + y 2 2 tht lies in the first qudrnt x, y. Therefore the surfce re Jcobin is d JdA + g x) 2 + g y) 2 da + y/) 2 + x/) 2 da The surfce integrl is then trnsformed to the double integrl over nd the ltter is evluted in polr coordintes: π2 2 d JdA + u2 udu π2 4 dθ + r/) 2 rdr π 2 ) z dz 2 2 where the substitutions u r/ [, ], dr du, nd z + u 2 [, 2], dz 2udu, hve been mde. 2. Find the center of mss of the prt of the sphericl shell 2 x 2 + y 2 + z 2 b 2 tht lies bove the xy plne if the mss density t point is proportionl to the distnce from the point to the z xis. olution: ince the shpe of the shell is invrint under rottions bout the z xis, while the mss density depends only on the distnce from the z xis nd, hence, is lso invrint under such rottions, the center of mss must be on the z xis, x c y c. The mss density is σ k x 2 + y 2 for some constnt k. Therefore, using sphericl coordintes to evlute the totl mss M nd the moment M z, σ kρsinφ) nd dv ρ 2 sinφ)dv, M π σ dv k sin 2 φ)dφ π2 k 8 b4 4 ) M z zσ dv k z c 5 b5 5 ) M z M 6 5π b b ρ 3 dρ πk π b sin 2 φ)cosφ)dφ b 5 5 b 4 4 b ρsinφ) ρ 2 sinφ)dρdφdθ 6 cos2φ))dφ 4 b4 4 ) ρcosφ)ρsinφ) ρ 2 sinφ)dρdφdθ ρ 4 dρ 3 sin3 φ) π/2 5 b5 5 ) 3. vlute the line integrl of the vector field F zx,yz, z 2 long the portion of the helix rt) 2sin t, 2cos t, t tht lies inside the ellipsoid x 2 + y 2 + 2z 2 6.
olution: The vlues of the prmeter t corresponding to the points of intersection of the helix with the ellipsoid determined by the substitution of the prmetric equtions into the ellipsoid eqution: 4sin 2 t) + 4cos 2 t) + 2t 2 6 4 + 2t 2 6 t ± o, t for the prt of the helix tht lies inside the ellipsoid. Then the substitution of the prmetric eqution into F gives the vector field Ft) on the helix nd the line integrl is reduced to n ordinry integrl Ft) r t)dt 2t sint), 2t cost), t 2 2cost), 2sint), dt 4t cost)sint) 4t cost)sint) + t 2 ) dt t 2 dt 2 3 4. etermine whether the vector field F 2x yz, 4y xz, 6z xy is conservtive nd evlute its line integrl long the contour tht consists of the four stright line segments: A,, ) B, 2, 3),, ), 2, 3),, ). olution: The components of the vector re polynomil nd, hence, the domin of F is the whole spce which is simply connected. The curl of F vnishes: ê ê 2 ê 3 F det x y z x x), y y)), z z) 2x yz 4y xz 6z xy The vector field is conservtive nd there exists fx, y, z) such tht F grd f. By the fundmentl theorem for line integrls f) FA) f,, ) f,, ) The problem is reduced to finding the potentil f. Integrting the reltion f x F with respect to x, fx, y, z) x 2 xyz + gy, z) for some gy, z). The substitution of f into the reltion f y F 2 yields g y 4y from which it follows tht gy, z) 2y2 + hz) for some hz). Finlly, the substitution of f into f z F 3 yields the eqution for h: h 6z or hz) 3z 2 + so tht fx, y, z) x 2 + 2y 2 + 3z 2 xyz +. Therefore f,, ) f,, ) 5. Alterntively: Owing to pth-independence of the line integrl of conservtive vector field, ny convenient pth from A to cn be chosen. For exmple, stright line: x t, y t, z t, t [, ]. The line integrl is then evluted just like in Problem 3: the function Ft) r t) 2t 3t 2 is integrted over [, ] the integrl is equl to 5). 5. Find the flux of the vector field F zx, zy, xy cross the prt of the sddle surfce z xy tht lies between two cylinders x 2 + y 2 nd x 2 + y 2 4. The surfce of integrtion is oriented upwrd s viewed from the tip of the z xis. olution: The surfce is the grph z gx, y) xy where x, y) nd is the nnulus x 2 + y 2 2 with rdii nd 2. An upwrd norml is n g x, g y, y, x,. The surfce intgerl for the flux is converted into double integrl by the substitution d n da
nd ˆn n/ n ) nd the ltter is evluted in polr coordintes: F ˆnd 2 F n da zx, zy, xy y, x, da zgx,y) zxy π 2x 2 y 2 + xy)da 2 x 2 y 2 da 2 π sin 2 2θ)dθ 6 26 ) 2 8 π sin 2 θ)cos 2 θ)dθ r 5 dr cos4θ))dθ 2π 4 Note tht the integrl of xy over vnishes by symmetry the first equlity in the second eqution line bove). 6. Use the contour trnsformtion in the line integrl bsed on the Green s theorem to evlute the line integrl of the vector field F e x siny) 2x, e x cosy) 2 over the upper prt of the circle x 2 + y 2 2x y ) oriented from the origin to the point 2, ). Hint: Apply Green s theorem to the line integrl of F long the boundry of the hlf-disk x 2 + y 2 2x, y. olution: Let be the contour of integrtion in question nd 2 be the intervl x 2, y, oriented from, ) to, 2). Then the union 2 ) is the positively counterclockwise)oriented boundry of the hlf-disk described in the hint. By Green s theorem F dx + F 2 dy + 2 ) F2 F dx + F 2 dy x F ) da y e x cosy) e x cosy))da becuse is the hlf-disk of rdius centered t, )). Therefore F dx + F 2 dy F dx + F 2 dy F dx + F 2 dy 2 Note tht dy nd y on 2. F x, )dx 2x)dx 4 7. Use Green s theorem to find the re of n stroid bounded by the curve x cos 3 t), y b sin 3 t), t 2π. olution: One hs A) xdy 3 π 6 b 3 π 6 b 3π 8 b π π xt)dyt) 3b cos 4 t)sin 2 t)dt 3 π 4 b cos 2 t)sin 2 2t)dt ) ) + cos2t) cos4t) dt cos4t) + cos2t) 2[ ] ) cos2t) + cos6t) dt where the double ngle formuls were used sin2t) 2cost)sint), sin 2 u) cos2u)), nd 2 cos 2 u) + cos2u)), nd the product formul cosα)cosβ) [cosα β) + cosα + β)] to 2 2 evlute the integrl.
8. Use tokes theorem to evlute the line integrl of the vector field F y, x z, y long the closed contour tht is the intersection of the cylinder x 2 + y 2 nd the prboloid z x ) 2 + y 2 ; the contour is oriented clockwise s viewed from the tip of the z xis. olution: hoose the grph z gx, y) x ) 2 + y 2, where x, y) nd is the disk x 2 + y 2, to be the surfce whose boundry is the contour of integrtion. The induced orienttion of is downwrd in order for to hold nd norml vector is n g x, g y, 2x ), 2y,. Next Then by tokes theorem, ê ê 2 ê 3 B curlf det x y z 2,, 2 y x z y B ˆnd B n da 2,, 2 2x ), 2y, da zgx,y) 4x 6)dA 6 da 6A) 6π Note tht the integrl of x over vnishes by symmetry. Alterntively: One cn show tht lies in plne whose prt encircled by cn be chosen s in the tokes theorem. Indeed, the substitution of y 2 x 2 into z x ) 2 + y 2 yields z x 2 2x++ x 2 ) 2 2x which implies tht the coordintes of the points of intersection i.e., points of ) lie in the plne z 2 2x so tht the grph z gx, y) 2 2x, x, y), cn lso be used. 9. Use the Guss-Ostrogrdsky divergence) theorem to evlute the flux of the vector field F 3xy 2 z + e zy, 3yx 2 z, z 3 + cosxy) cross the closed surfce oriented outwrd tht is the boundry of the portion of the bll x 2 + y 2 + z 2 R 2 in the first octnt x, y, z ). Wht kind of source does the field hve inside sink or fucet)? olution: By the divergence theorem the flux surfce integrl cn be reduced to triple integrl over the solid enclosed by the surfce of integrtion, which is evluted in sphericl coordintes: F ˆn d 3 divfdv R 3 π 2 6 R6 π 4 R6 4 sin4 φ) π/2 π 6 R6 + π R5 > 3x 2 z + 3y 2 z + 3z 2 )dv ρ 3 sin 2 φ)cosφ) + ρ 2 cos 2 φ) ) ρ 2 sinφ)dρdφdθ sin 3 φ)cosφ)dφ + 3 π 2 5 R5 cos 2 φ)sinφ)dφ 3π R5 3 cos3 φ) π/2 ince the outwrd net flux is positive, the vector field F hs net fucet source in the solid enclosed by.
xtr credit. Let be simple, smooth, closed curve in the plne n r d, where r x, y, z, nd let A be the re of the prt of the plne encircled by. If is oriented counterclockwise s viewed from the tip of the norml vector n, find the line integrl of the vector field F n r long. xpress your nswer in terms of the given quntities A, d, nd n. olution: Using the bc cb rule for double cross product, the curl of F is computed: B curlf n r) n r) n )r ndivr n ) x, y, z 3n n 2n Note n )x n nd similrly for the other components of r. By tokes theorem B ˆnd 2n ˆnd 2 n d 2 n A becuse ˆn n/ b the unit norml vector). Remrk. The curl of F cn lso be clculted by direct evlution of the cross products using the determinnts): F n 2 z n 3 y, n 3 x n z, n y n 2 x so tht ê ê 2 ê 3 B curlf det x y z 2n, 2n 2, 2n 3 2n n 2 z n 3 y n 3 x n z n y n 2 x