alculus III - Problem Drill 4: tokes and Divergence Theorem Question No. 1 of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the 1. Use tokes' Theorem to evaluate where F(x, y, z) = (y)i (xz)j + (yz )k and is the surface of the paraboloid z = x + y bounded by z = and is the curve of intersection of the plane z = with the paraboloid. Question #1 hoice =. =. () = 4. (D) =. (E) = 1. A. orrect! Well done, you have correctly applied tokes Theorem that uses a double integral to evaluate the line integral required. In this case you have correctly used the downward unit normal. B. Incorrect! Almost, this answer indicates that you used the incorrect orientation for the curve and therefore the upward unit normal. This has produced a negative result. Redo the problem using the downward unit normal so you get a positive orientation.. Incorrect! This answer indicates that you incorrectly simplified the integral. Most likely you missed a division by when substituting for z. Redo the simplification. D. Incorrect! This answer indicates that you missed the r in the conversion to polar coordinates which involves da = rdrdθ. The r needs to be multiplied through before integrating. E. Incorrect! This result suggests that you have incorrectly applied a trigonometric identity when evaluating the final integral of the double integral. Recheck the identity and re-evaluate the integral. We begin by sketching the curve to get a sense of what we need to find. olution As shown on the graph the curve is the circle x + y = 4. To apply tokes' Theorem we first compute curl F= = ( z + x) i ( ) j+ ( x ) k = ( z + x) i ( z+ ) k y xz yz By orientating downwards we have with a positive orientation. The projection P of onto the xyplane is the disk x + y 4. Hence, we let z = g(x, y) = ½( x + y ) and therefore g ( x, y) = x and g ( x, y) = y x and the downward normal vector N is N= gx( x, y) i+ gy( x, y) j k. Now by tokes' Theorem (and using polar coordinates) F i dr = curl F i N d = z + x i z+ k i xi + yj k dxdy olution ( ) ( ) ( ) ( ) y P x + y x + y = ( xz + x + z + ) dxdy = x + x + + dxdy P P 4 7 4 4 cos r cos r r cos r cos r r 4 8 4 8 = cosθ 1 ( cosθ) 8dθ sinθ θ sin θ 8θ + + + = + + + = 7 7 ( ) = r θ + r θ + + rdrdθ = θ + θ + + dθ = RapidLearningenter.com Rapid Learning Inc. All Rights Reserved
Question No. of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the. Use tokes' Theorem to evaluate if F(x, y, z) = (e x )i + (e x )j + (e z )k where is the boundary of the part of the plane x + y + z = in the first octant. is oriented counterclockwise when viewed from above. Question # hoice = i () = e+ 4 (D) = e 4 (E) = e 4 e 4 F dr = e A. Incorrect! This result suggests that you have not used the correct limits of integration when integrating the final integral for the double integral. Be sure you have not reversed the limits of integration for the double integral. B. Incorrect! This result suggests that you have incorrectly determined the limits of integration for the first integral of the double integral. It appears you have used instead of the line x for the upper limit of integration. Recheck that you have the correct path and projection then re-evaluate the integral.. Incorrect! Almost. This result suggests that you incorrectly applied the Fundamental Theorem of alculus or made a simplification error in the evaluation of the final integral. heck for both these cases and re-evaluate the integral. D. Incorrect! Almost. This result suggests that you incorrectly applied the Fundamental Theorem of alculus or made a simplification error in the evaluation of the final integral. heck for both these cases and re-evaluate the integral. E. orrect! Well done, you have correctly applied tokes Theorem by using a double integral to evaluate the line integral required. In this case you have correctly used the upward unit normal to the plane surface. We begin by sketching the curve to get a sense of what we need to find. olution As shown on the graph the curve encloses a planar region so we can take a portion of this region in the first octant to be the surface. That is, the portion of the plane x + y + z =. Also, since is oriented counterclockwise, we orientate upward. We let z = g(x, y) = ½( x y) and therefore g x (x, y) = x and g y (x, y) = ½y. By orienting upwards the upward normal vector N is 1 1 N= gx( x, y) i gy( x, y) j+ k = ( x) i ( y) j+ k = xi+ yj+ k. To apply tokes' Theorem we first compute x x curl F= = ( ) i ( ) j+ ( e ) k = ( e ) k x x z e e e Therefore, by tokes Theorem with the projection P of onto the xy-plane being the triangle with vertices (1,, ), (,, ), and (,, ) we have 1 1 x x x x F i dr = ( curl F i N) d = ( e ) k i xi + yj + k da = ( e ) da = ( e ) dydx P P 1 x 1 x 1 1 1 1 x x x x x x x ( ) ( ) ( ) ( ) ( ) = e dydx = ye dx = x e dx = e xe dx = e dx xe dx By using integration by parts on the last integral we obtain olution: 1 1 x x x 1 x x 1 1 1 1 ( ) ( ) ( ) (() ) ( ) e dx xe dx = e xe e = e e 1 e e e = e 4 = e 4 RapidLearningenter.com Rapid Learning Inc. All Rights Reserved
Question No. of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the. Let be the rectangle in the plane z = y orientated as shown in the figure below and let F(x, y, z) = (x )i + (4xy )j + (y x)k. Find: Question # hoice olution F i d r = 7 F i d r = 4 () F i d r = 9 (D) F i d r = 9 1 (E) F i d r = A. Incorrect! This result suggests that you have used the upward normal vector rather that the downward normal vector. The orientation displayed by the figure requires the downward normal vector be used for a positive orientation. Re-evaluate the integral using the downward normal vector. B. Incorrect! This result suggests that you have not used the correct limits of integration when integrating the final integral from the double integral. Be sure you have not reversed the limits of integration for the double integral.. orrect! You have correctly applied tokes Theorem by using a double integral to evaluate the line integral required. In this case you have correctly used the downward unit normal to the plane surface. D. Incorrect! This result suggests that the negative sign in the double integral has been lost in the integration process. Go back and check the simplification and evaluation of the integral to find where the negative sign was missed. E. Incorrect! This result suggests that the incorrect partial derivatives where obtained and therefore the downward normal is incorrect. This error then carried over into an incorrect dot product. Go back and reevaluate the partial derivatives for z = g(x, y) = y and then find the downward normal again. To evaluate the integral directly would require four separate integrations, one integration over each side of the rectangle. onsequently, it is more efficient to use tokes Theorem which will require the evaluation of only one surface integral over the rectangular surface bounded by. For the positive direction of to be as shown in the figure above, the surface must be oriented by downward normals. ince z = g(x, y) = y then g x (x, y) = and g y (x, y) = 1. The downward normal vector N becomes N= gx( x, y) i+ gy( x, y) j k = ( ) i+ ( 1) j k = j k. To apply tokes' Theorem we first compute curl F= = ( xy ) i ( y ) j+ ( 4y ) k = ( xy ) i ( y ) j+ ( 4y ) k x 4xy y x Therefore, by tokes Theorem with the projection P of onto the xy-plane being the triangle with vertices (1,, ), (,, ), (,, ) and (1,, ) we have 4 4 F i dr = curl F i N d = xy i y j + y k i j k da= y y da ( ) ( ( ) ( ) ( ) ) ( ) ( ) P P 1 1 1 1 4 ( ) y [ ] = y 4y dydx = + y dx = 9 + 81 dx = 9 dx = 9 olution F i d r = 9 RapidLearningenter.com Rapid Learning Inc. All Rights Reserved
Question No. 4 of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the 4. A liquid is swirling around a cylindrical container of radius. The motion of the liquid is described by the velocity field F( xyz,, ) = ( xy ) i+ j+ zk Find the approximate rate of circulation of the liquid on the upper surface of the cylindrical container and then evaluate curl F i N d Question #4 hoice ( curl F i N) d = 8 ( curl F i N) d = 8 () ( curl F i N) d = 16 (D) ( curl F i N) d = 16 (E) ( curl F i N) d = ( ) A. Incorrect! Almost, but this answer indicates that there has been an error made in the simplification of the integrals or their evaluation. A negative sign was produce by curl F but was lost during the integration process. Recheck curl F and check through the integration process again. B. orrect! Outstanding, you have correctly determined the approximate rate of circulation that has lead to the correct value of the tendency of the fluid to circulate around the boundary, namely 8.. Incorrect! Almost, but this answer indicates that there has been an error made in the simplification of the last integral when applying a trigonometric identity. Recheck the constants in the simplification. D. Incorrect! This result suggests that that there has been an error made in the simplification of the last integral when applying a trigonometric identity. Recheck the constants in the simplification plus there is a sign out of place. Possibly lost in the integration process. Recheck the integration process. E. Incorrect! This result suggests that you have incorrectly determined curl F. Recheck your calculation of curl F and also the application of the Fundamental Theorem of alculus in the integration process. We will first sketch a picture to get a sense of what we are required to find. Also, since the approximate rate of circulation is given by (curl F. N), we will need to find curl F. olution To calculate curl F we use curl F = = i j+ ( xy ) k = ( xy ) k xy 1 z The approximate rate of circulation is given by (curl F. N). Letting N = k we obtain curl F i N = x y k i k = x y We now evaluate ( ) ( ) ( curl F i N) That is, since the surface is projected onto the circular region x + y 4 we have curl F i N d = x y da d ( ) ( ) We convert this to polar coordinates using x = rcos θ, y = rsin θ, and da = rdrdθ. That is, ( ) ( ) ( )( curl F i N = = cos sin ) = ( sin cos ) P P d x y da r θ r θ rdrdθ r θ θ drdθ 6 r = sin θ cos θ dθ ( sin θcos θ) dθ 8 ( sin θ) dθ 4 ( 1 cos4θ) dθ 6 = = = sin4θ = 4 θ = 4 ( ) ( ) = 4( ) = 8 4 olution curl F N d = ( i ) 8 RapidLearningenter.com Rapid Learning Inc. All Rights Reserved
Question No. of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the. Use tokes Theorem to calculate the flux of the curl of the vector field F(x, y, z) = (z)i + (x) j + (y)k across the surface in the direction of the outward normal vector when is given in the parametric form r(u, v) = (ucos v)i + (usinv)j + (4 u )k where u and v. Question # hoice olution ( curl F i N) d = ( curl F i N) d = 1 () ( curl F i N) d = 1 (D) ( curl F i N) d = 6 (E) 64 ( curl F i N) d = 1 A. Incorrect! This answer suggests that you have made a mistake either in curl F or in the simplification of the cross-product r u X r v. Recheck these calculations and the simplification process. B. Incorrect! This answer indicates that you have either not found the outward normal vector or you have made an algebraic mistake when applying the Fundamental Theorem of alculus to the integral.. orrect! Outstanding, you have correctly determined the flux of the curl of the vector field F(x, y, z) = (z)i + (x)j + (y)k using tokes Theorem when the surface is in parametric form. D. Incorrect! This result suggests that you have made an algebraic error in the evaluation of the integral. Recheck the algebra and the arithmetic used to evaluate the integrals. E. Incorrect! This result suggests that you have made a mistake in applying the Fundamental Theorem of alculus when evaluating the last integral. Re-evaluate the last integral and be careful of the signs. We first compute curl F. That is, curl F= = i ( ) j+ k = i+ j+ k z x y Next, since the surface is in parametric form we need to calculate r u and r v. That is, ru = ( ucosv) i+ ( usinv) cosvj+ ( 4 u ) k = ( cosv) i+ ( sinv) j+ ( u) k u u u and rv = ( ucosv) i+ ( usinv) cosvj+ ( 4 u ) k = ( usinv) i+ ( ucosv) j+ ( ) k v v v We can now find the normal vector r u X r v. That is, ru rv = cosv sinv u = ( u cosv) i ( u sinv) j+ ( ucos v + usin v) k = ( u cosv) i+ ( u sinv) j+ ( u) k usinv ucosv Therefore, r u X r v becomes (( ) ( ) ( ) ) ( ) ru rv = u cosv + u sinv + u = u cos v + sin v + u = u u + and the unit normal N is and therefore ( curl F i N) ( ) Hence, 4 4 4 1 ru rv N = and d = ru rv dudv r r u v (( u cosv) i+ ( u sinv) j+ ( u) k) ( u 4u + 1) ( ) d = + + i u 4u + 1 dudv ( ) i (( cos ) i ( sin ) j ( ) k) ( 1 cos 4 sin ) = + + u v + u v + u dudv = u v + u v + u dudv 1 4 ( F i N) ( 1 4 ) olution curl d = u cosv + u sinv + u dudv = u cosv + u sinv + u dv 8 8 = cosv sinv 6 dv sinv cosv 6v 1 1 + + = + = + + = ( curl F i N) d = 1 RapidLearningenter.com Rapid Learning Inc. All Rights Reserved
Question No. 6 of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the 6. Find the divergence of the vector field defined by F = x i + y j + z k and determine whether the flow field F as defined is free of sources and sinks. If not, find the location of all sources and sinks. Question #6 () (D) (E) div F= x i+ y j+ z k ; sources at all points except the origin; no sinks div F = x + y + z ; sink at the origin; no sources div F = x + y + z ; incompressible at the origin; no sinks div F = x + y + z ; sinks at all points except the origin; no sources div F = x + y + z ; sources at all points except the origin; no sinks hoice olution A. Incorrect! This answer suggests you have failed understand the definition of div F. Remember, M N Q div F = i F= + + which means that since it is a dot product then a scalar must be the result not a vector. Recalculate div F. You have correctly classified the points. B. Incorrect! You have correctly determined div F but misclassified the type of points. Remember, if div F > then the point is a source. In this case, div F > for all points (x, y, z) except (x, y, z) = (,, ) where div F =. Review the classification conditions.. Incorrect! You have correctly determined div F but misclassified the type of points. Remember, if div F > then the point is a source. In this case, div F > for all points (x, y, z) except (x, y, z) = (,, ) where div F =. Review the classification conditions. D. Incorrect! Almost, you have correctly determined div F but misclassified the type of points. Remember, if div F > then the point is a source. In this case, div F > for all points (x, y, z) except (x, y, z) = (,, ) where div F =. Review the classification conditions. E. orrect! Well done, you have correctly found div F and correctly classified the points as all sources except at (,, ) using div F > for all points (x, y, z) except (x, y, z) = (,, ) where div F =. The divergence of F is calculated for a vector field defined by F= Mxyz (,, ) i+ Nxyz (,, ) j+ Qxyz (,, ) k using M N Q div F = i F = + + Hence, That is, div F = i F = x + y + z = x + y + z ( ) ( ) ( ) div F = i F = x + y + z A point (x, y, z ) in a vector field can be classified as a source, a sink, or incompressible by the following: 1. ource, if div F >.. ink, if div F <.. Incompressible, if div F =. onsequently, since div F = x + y + z > for all points (x, y, z) except (x, y, z) = (,, ) we conclude that there are sources at all points except the origin which is incompressible. There are no sinks. olution div F = x + y + z ; sources at all points except the origin; no sinks RapidLearningenter.com Rapid Learning Inc. All Rights Reserved
Question No. 7 of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the 7. Use the Divergence Theorem to calculate the surface integral F i N d ( ) That is, the flux of F across where is the surface of the cube with vertices (±1, ±1, ±1) and F = y z i + 9x yz j 4xy k Question #7 hoice ( F i N) d = 1 ( F i N) d = 7 () ( F i N) d = 1 (D) ( F i N) d = 8 (E) ( F i N) d = A. Incorrect! This answer indicates that you have integrated only half the required region. The limits of integration should be between 1 and 1 for each integral. You appear to have only used to 1 for the limits of integration. Recalculate the integrals using the correct limits of integration. B. Incorrect! This answer suggests you have made the same mistake in each integral evaluation. Remember to add one to the exponent and divide the term by the new exponent each time you integrate the variable using the power rule.. Incorrect! Almost. This result suggests that you have made an error in the evaluation of the last integral. Remember to add one to the exponent and divide the term by the new exponent. D. orrect! Outstanding! You have correctly determined the surface integral using the Divergence Theorem which shows that the surface integral can be evaluate using one triple integral. E. Incorrect! This answer suggests that you have incorrectly determined div F. Recalculate div F as all the partial derivatives are not zero. We first sketch the surface to get a sense of what we want to find. That is, olution We now find the divergence of the vector field. That is, div F = ( y z) + ( 9xyz) + ( 4xy ) = + 9xz + = 9xz onsequently, if R denotes the solid cube enclosed by, then it follows from the Divergence Theorem that F i N d = div F dv = 9x z dv ( ) ( ) ( ) R R 1 1 1 = ( ) 1 1 1 9x z dxdydz ( () ) ( ) ( ) 1 1 1 1 1 = x z dydz 1 z 1 z dydz 1 1 = 1 1 1 1 1 1 1 ( 6z ) dydz 6yz dz 1 1 1 1 1 ( () ) ( ( ) ) 1 61 6 1 ( 1 ) = = = z z dz z dz 1 = 1 ( () ( )) z 1 1 = 4 = 4 1 4 1 = 4+ 4 = 8 olution ( F i N) d = 8 RapidLearningenter.com Rapid Learning Inc. All Rights Reserved
Question No. 8 of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the 8. Use the Divergence Theorem to calculate the surface integral F i N d ( ) That is, the flux of F = x i + xz j + y zk across where is the surface of the solid bounded by the paraboloid z = 4 x y and the xy-plane. Question #8 hoice ( F i N) d = 64 ( F i N) d = () ( F i N) d = 88 (D) ( F i N) d = 4 (E) ( F i N) d = A. orrect! Well done! You have correctly determined the surface integral using the Divergence Theorem which shows that the surface integral can be evaluate using one triple integral. B. Incorrect! This answer indicates that you have missed carrying the constant that came from the calculation of div F through the integration process. Go back and check div F and ensure that the constant carries though the entire integration process when evaluating the triple integral.. Incorrect! This answer suggests you have incorrectly integrated with respect to r in the second integral of the triple integral. Remember that the power rule states that one is added to the exponent and then you must divide the term by the new exponent value. Recheck your integration. D. Incorrect! This answer suggests that you have either incorrectly executed or omitted the integration step that required the integration with respect to z. Recheck the integration process for evaluating the triple integral. E. Incorrect! This answer suggests that you have either incorrectly determined div F or made an error when applying the Fundamental Theorem of alculus in the process of evaluating the integral. Recalculate div F since all the partial derivatives are not zero. Also, check the integration process. We first sketch the surface to get a sense of what we want to find. That is, olution We now find the divergence of the vector field. That is, div F = x + xz + y z = x + + y = x + y ( ) ( ) ( ) ( ) onsequently, if R denotes the solid paraboloid enclosed by, then it follows from the Divergence Theorem that F i N d = div F dv = x + y dv ( ) ( ) ( ) R R We now convert this integral to an integral in cylindrical coordinates. That is, olution 4 r ( F i N ) = ( F) = ( + ) = ( ) d div dv x y dv r rdzdrdθ R R 4 r 4 ( ) r ( ) = r dzdrdθ = r z drdθ r 4 r = drdθ 6 4 r = ( 4r r ) drdθ = r dθ = 16 dθ = 16 dθ 6 [ θ ] = 16 = ( F i N) d = RapidLearningenter.com Rapid Learning Inc. All Rights Reserved
Question No. 9 of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the 9. Use the Divergence Theorem to find the outward flux of F = x i + y j + z k across the boundary of the solid sphere x + y + z 4. Question #9 hoice 19 The outward flux = 84 The outward flux = () 18 The outward flux = (D) The outward flux = (E) 84 The outward flux = A. Incorrect! This result suggests that an error has been made in the second integral of the triple integral. The integral of sin φ is cos φ but the integral is to be evaluated between and. When applying Fundamental Theorem of alculus this come out to not 1. Re-evaluate the second integral. B. orrect! Well done, you have correctly determined the outward flux using the correct formula.. Incorrect! This answer indicates that you have missed carrying the constant that came from the calculation of div F through the integration process. Go back and check div F and ensure that the constant carries though the entire integration process of evaluating the triple integral. D. Incorrect! This answer suggests that you have either incorrectly determined div F or made an error when applying the Fundamental Theorem of alculus in the process of evaluating the integral. Recalculate div F since all the partial derivatives are not zero and the integration process. E. Incorrect! This answer suggests that you have incorrectly evaluated the second integral of the triple integral by using the incorrect limits of integration. The variable φ should be integrated between and but it appears you have integrated between and. Re-evaluate the second integral using the correct limits of integration. We first sketch the surface to get a sense of what we want to find. That is, olution The outward flux is found using the Divergence Theorem by evaluating F i N d = div F dv ( ) ( ) onsequently, we determine div F first. That is, div F = x + y + z = x + y + z = x + y + z R ( ) ( ) ( ) ( ) onsequently, if R denotes the solid sphere enclosed by, then it follows from the Divergence Theorem that F i N d = div F dv = x + y + z dv ( ) ( ) ( ) R R We now convert this integral to an integral in spherical coordinates. That is, F i N d = div F dv = x + y + z dv = sin d d d ( ) ( ) ( ) ( ) R R ρ ρ φ ρ φ θ ρ 96 96 = sinφ dφdθ [ sinφ] dφdθ [ cosφ] = = dθ 96 96 96 19 84 = [ ] [ ] ( ) [ ] cosφ dθ = cos cos dθ dθ θ + = = = olution 84 The outward flux = RapidLearningenter.com Rapid Learning Inc. All Rights Reserved
Question No. 1 of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the 1. Use the Divergence Theorem to find the outward flux of F = yj + xyj zk across the region inside the solid cylinder x + y 4 between the plane z = and the paraboloid z = x + y. Question #1 The outward flux = 4 The outward flux = 8 () The outward flux = 8 (D) The outward flux = 4 (E) The outward flux = hoice A. Incorrect! This answer suggests that you have incorrectly evaluated the third integral of the triple integral by using the incorrect limits of integration and mixing up the sign. The variable θ should be integrated between and but it appears you have integrated between and. Re-evaluate the third integral using the correct limits of integration and recheck your signs. B. Incorrect! This answer suggests that you have made an error in the evaluation of the final integral when applying the Fundamental Theorem of alculus. Re-evaluate the last integral and be careful of the signs.. orrect! Outstanding! You have correctly determined the outward flux using the correct formula. D. Incorrect! This result suggests that you have incorrectly converted to cylindrical coordinates. Remember, dv = rdzdrdθ. It appears you have omitted the r when converting to cylindrical coordinates. Re-convert to cylindrical coordinates and re-evaluate the triple integral. E. Incorrect! This answer suggests that you have either incorrectly determined div F or made an error when applying the Fundamental Theorem of alculus in the process of evaluating the integral. Recalculate div F as all the partial derivatives are not zero and the integration process. We first sketch the surface to get a sense of what we want to find. That is, olution The outward flux is found using the Divergence Theorem by evaluating F i N d = div F dv ( ) ( ) R onsequently, we determine div F first. That is, div F = ( y) + ( xy) + ( z) = + x 1= x 1 onsequently, if R denotes the solid cylinder enclosed by, then it follows from the Divergence Theorem that F i N d = div F dv = x 1 dv ( ) ( ) ( ) R R We now convert this integral to an integral in cylindrical coordinates. That is, since z = x + y, we have z = r and r F i N d = div F dv = x 1dV = r cosθ 1rdzdrdθ olution ( ) ( ) ( ) ( ) R R r ( cos ) r cos = r θ r dzdrdθ = zr θ zr drdθ 4 4 r r = r cosθ r drdθ = cosθ dθ = cosθ 4 dθ 4 = sinθ 4θ sin( ) 4( ) sin( ) 4( ) 8 = = The outward flux = 8 RapidLearningenter.com Rapid Learning Inc. All Rights Reserved