hw5_sol - Mathematics 20E - Spring 2009 - Homework 5 June...

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Unformatted text preview: Mathematics 20E - Spring 2009 - Homework 5 June 9, 2009 Please complete the following problems and turn them in to the homework drop box in the sixth floor of AP&M labelled "Math 20E / Ryan Szypowski" by Friday, June 5 at 5:30pm. Only a subset of the problems will be graded; leave some blank at your own risk. 1. In this problem, we will show that a particular vector field is path independant from first principles without invoking any theorems about conservative vector fields. Consider the constant vector field F = i + j + k. Let C1 and C2 be any two oriented curves with the same starting point and ending point. That is, let c1 : [a1 , b1 ] 3 and c2 : [a2 , b2 ] 3 be parameterizations of C1 and C2 respectively. Then, knowing only that c1 (a1 ) = c2 (a2 ) and c1 (b1 ) = c2 (b2 ), show F ds = C1 C2 F ds. 2. Recall that if F(x, y, z) = F1 (z, y, z)i + F2 (x, y, z)j + F3 (x, y, z)k satisfies F = 0 then F = G for some G. Prove that the following 1 construction of G works: G(x, y, z) = G1 (x, y, z)i + G2 (x, y, z)j + G3 (x, y, z)k z y G1 (x, y, z) = 0 F2 (x, y, t) dt - 0 z F3 (x, t, 0) dt G2 (x, y, z) = - 0 F1 (x, y, t) dt G3 (x, y, z) = 0. Note however, that this construction is clearly not unique since we could add the gradient of any function to G and still achieve this result. 3. First, show that the vector field F = (x2 + 1)i + (z - 2xy)j + yk satisfies F = 0. Then, use the above construction to find G such that F = G. Solution: Notice that F = 2 (x + 1) + (z - 2xy) + (y) x y z = 2x - 2x + 0 = 0. Also, F is a nice, smooth function everywhere. Thus, F = G for 2 some vector field G. In order to find, G, we use the above construction: z y G1 (x, y, z) = 0 z F2 (x, y, t) dt - 0 y F3 (x, t, 0) dt t dt = 0 t - 2xy dt - 0 = 1 1 2 z - 2xyz - y 2 , 2 2 z G2 (x, y, z) = - 0 z F1 (x, y, t) dt x2 + 1 dt 0 = - = -x2 z - z G3 (x, y, z) = 0. 4. Determine which of the following vector fields is the gradient of some scalar field, and if it is, find the scalar field: (a) F(x, y, z) = xi + yj + zk. Solution: In order to determine if F is the gradient of some vector field, we can simply compute F and see if it is 0. Here, i F = = 0. Thus, F is the gradient of some scalar field. To find it, we use the construction x y z x j y k z x y z f (x, y, z) = 0 x F1 (t, 0, 0) dt + 0 y F2 (x, t, 0) dt + 0 z F3 (x, y, t) dt = 0 t dt + 0 t dt + 0 t dt = 1 2 (x + y 2 + z 2 ). 2 3 (b) F(x, y, z) = xyz(i + j + k). Solution: In order to determine if F is the gradient of some vector field, we can simply compute F and see if it is 0. Here, i F = x j y k z xyz xyz xyz = (xz - xy)i + (xy - yz)j + (yz - xz)k. Thus, F is not the gradient of some scalar field. (c) F(x, y, z) = (x2 + y 2 + z 2 )i + 2xyj + 2xzk. Solution: In order to determine if F is the gradient of some vector field, we can simply compute F and see if it is 0. Here, i F = x 2 j y k z x2 + y + z 2 2xy 2xz = 0. Thus, F is the gradient of some scalar field. To find it, we use the construction x y z f (x, y, z) = 0 x F1 (t, 0, 0) dt + 0 y F2 (x, t, 0) dt + 0 z F3 (x, y, t) dt = 0 t2 dt + 0 2xt dt + 0 2xt dt = 1 3 x + xy 2 + xz 2 . 3 5. Verify Gauss' theorem by computing both F dV V 4 and F dS V where V = [0, 1] [0, 1] [0, 1] and F = x2 i + y 2j + z 2 k. It is probably easiest to compute the surface integral by recognizing the integral over each of the 6 faces as some particular area integral. Solution: For the first part, we simply compute 1 1 0 1 1 0 1 F dV V = 0 2x + 2y + 2z dx dy dz x2 + 2xy + 2xz 0 1 0 1 1 x=0 = = 0 1 0 dy dz 1 + 2y + 2z dy dz y + y 2 + 2yz 0 1 1 y=0 = = 0 dz 2 + 2z dz 1 z=0 = 2z + z 2 = 3. For the scond, we split V up into 6 faces, and compute the surface integral over each. Here, I'll show the details for 2 faces: the face F1 = {(x, y, z) : x = 0, 0 y, z 1} and F2 = {(x, y, z) : x = 1, 0 y, z 1}. For F1 , recognise that the unit outward normal is n = (-1, 0, 0). Thus, (x2 , y 2, z 2 ) dS = F1 F1 (x2 , y 2, z 2 ) ndS -x2 dS. F1 = 5 However, x = 0 on F1 , and so this integral will simply be 0. For F2 , recognise that the outward normal is n = (1, 0, 0). Thus, (x2 , y 2, z 2 ) dS = F2 F2 (x2 , y 2, z 2 ) ndS x2 dS. F2 = On F2 , x = 1, and so this will simply compute the surface area of F2 which is 1. The other faces can be done similarly and we arrive at a total integral of 3. 6. Compute the surface integral of F = x3 i + y 3 j + z 3 k over the surface of the unit sphere, S. That is, find F dS. S Solution: Let V be the unit ball whose boundary is S, the unit sphere. We now use Gauss' theorem to translate this into F dS = S V F dV 3x2 + 3y 2 + 3z 2 dV. V = This integral can be tackled easily by using spherical coordinates, 2 0 2 0 0 1 3x + 3y + 3z dV V 2 2 2 = 0 3r 4 sin dr d d 3 sin d d 5 = 0 = 0 6 sin d 5 12 = . 5 6 7. Let W = [0, 1] [0, 1] [0, 1] and F = (x + y + z)i + (y + z 2 )j + zk. Compute F dS. W Solution: We use Gauss' theorem and compute F dS = W W 1 F dV 1 0 0 1 = 0 3 dV = 3. 7 ...
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This note was uploaded on 10/22/2009 for the course CHEM 140A taught by Professor Whiteshell during the Spring '04 term at UCSD.

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