NAME:Mathematics 10B–010, Fall 2009, Examination 3Answer Key11. [15 points] Given the surface with parametrization X(u, v) = (u + v, u − v, v),find a function f(x, y) such that the surface satisfies the equation z = f(x, y). [Hint: Thesurface is a plane.]SOLUTIONThe parametrization yields x = u + v, y = u − v and z = v. We already have zexpressed in terms of u and v, and if we substitute z = v into the second equation we findthat y = u − z, so that u = y + z. We now have u and v in terms of y and z, and if wesubstitute this into the first equation we find that x = u + v = y + 2z, or equivalentlyx − y − 2z = 0.22. [15 points] Set up (but do not evaluate) an iterated integral formula for thearea of the surface z = exsin y where 0 ≤ x ≤ 1 and 0 ≤ y ≤ π. The integrand shouldbe expressed as a specific function, and all limits of integration should also be describedexplicitly.SOLUTIONLet f(x, y) = exsin y. Then the standard normal vector N(x, y) for the surface z =exsin y, which is the graph of f, is given by (−f1, −f2, 1), which is equal to(−exsin y, −excos y, 1)and its length is given by|N(x, y)| =p1 + e2x.The area of the surface is given by the integral of this function over the domain on whichthe function is defined, and therefore it is equal toZ10Zπ0p1 + e2xdy dx .33. [20 points] The parametrized curve Γ(t) = (cos t, sin t, cos 2t), where 0 ≤ t ≤ 2π,bounds the portion of the surface z = x2− y2which lies inside the cylinder x2+ y2≤ 1.Use Stokes’ Theorem to express the line integralZΓy3dx + 4xy2dy + ez2dzas a double integral over the planar region defined by x2+ y2≤ 1. You need not evaluatethis integral.SOLUTIONLet S be the surface which Γ bounds. The integrand in the line integral can be writtenas F · ds, where F(x, y, z) = (y3, 4xy2, ez2), and thus by Stokes’ Theorem the line integralequalsZ ZS(∇ × F) · S .If we use the standard 3 × 3 determinant formula for the curl, we see that∇ × F =Dyez2− Dz4xy2, Dzy3− Dxez2, Dx4xy2− Dyy3= (0, 0, y2) .Also, the standard upward normal is given by N(x, y) = (−zx, −zy, 1) = (−y, −x, 1), sothat (∇ × F) · N = y2and hence the surface integral is equal toZ ZDy2dy dxwhere D is the disk over which the parametrization is defined; namely D is the planarregion defined by x2+ y2≤ 1.This integral can be evaluated directly, but the problem does not require the evaluationof the integral.44. [20 points] Evaluate the surface integralR RSF · dS, where F(x, y, z) = (x, y, z)and S is the portion of the unit sphere x2+ y2+ z2= 1 which lies in the first octant. Takethe outward pointing normal to S.SOLUTIONWe shall use spherical coordinates to parametrize the surface asX(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ)where 0 ≤ θ, φ ≤12π. It follows immediately that F(θ, φ) = X(θ, φ), and since the surfaceis the unit sphere, the length of F is always 1.The standard normal vector to this parametrization is equal toN(θ, φ) = sin φ · (cos θ sin φ, sin θ sin φ, cos φ) = sin φ · Fand it follows thatF · N = sin φ|F|2= sin φ .Therefore the surface integral is equal toZπ/20Zπ/20sin φ dθ dφ =Zπ/20π2sin φ dθ dφ =π2− cos φπ/20=π2.55. [20 points] Let S be the surface z = exp(1 − x2− y2) where 0 ≤ x2+ y2≤ 1, andtake the upward pointing normal to S. If F is the vector field F(x, y, z) = (x, y, 3 − 2z),explain whyR RSF · dS is equal toR RTF · dT, where T is the surface z = 1 with theupward pointing normal , and as before 0 ≤ x2+ y2≤ 1. Using this, computeZ ZSF · dS .Note. If A is a real valued expression, then exp(A) = eA.SOLUTIONIf D is the region defined by 0 ≤ x2+ y2≤ 1 and 1 ≤ z ≤ exp(1 − x2− y2), then theboundary of S with the outward normal orientation is given by the union of S and −T ,where −T is equal to T with the downard pointing normal. Therefore we haveZ ZSF · dS −Z ZTF · dT =Z Z ZD∇ · F dV .Now ∇ · F = 1 + 1 − 2 = 0, and thereforeZ ZSF · dS =Z ZTF · dT .Thus it is only necessary to compute the right hand side. If we take the standardparametrization (x, y, 1) for T , the upward pointing normal N(x, y) is given by (0, 0, 1),so that F · N = 3 − 2z, which is equal to 1 since z = 1 on the surface T . Therefore thesurface integral over T is given by the ordinary integralZ ZD1 dA = Area(D) = π .66. [10 points] Suppose that S is a closed piecewise smooth surface bounding a regionD in coordinate 3-space, take the outward pointing normal for S, and let F be the vectorfield (2x, 3y, 4z). Find a constant k such thatVolume(D) =1kZ ZSF · dS .SOLUTIONThe divergence of the vector field is equal to 2 + 3 + 4 = 9, and therefore by theDivergence Theorem we have1kZ ZSF · dS =1kZ Z ZD∇ · 9 dV =9 · Volume(D)k=9k· volume(D) .We need the value of k such that 1 = 9/k, and this means that k =