Math 53 Worksheet Solutions - Flux and Divergence1. Is there a vector field G on R3such thatcurl G = hx sin y, cos y, z − xyi?Explain.Solution. No. For any differentiable vector field F, we havediv(curl(F)) = 0.However,div(hx sin y, cos y, z − xyi) = 1.(Check this.) As a result, the stated vector field cannot be the curl of a vector field G.2. Show that any vector field of the formF(x, y, z) = f (x)i + g(y)j + h(z)k,where f, g, and h are differentiable is irrotational.Solution. We must show curl(F ) = 0. Calculate the curl of F and observe that fdoes not depend on y or z, g does not depend on x or z, and h does not depend on xor y. As a result, the curl is zero.3. Show that any vector field of the formF(x, y, z) = f (y, z)i + g(x, z)j + h(x, y)k,is incompressible.Solution. We must show div(F ) = 0. Note thatdiv(F ) =∂∂xf(y, z) +∂∂yg(x, z) +∂∂zh(x, y) = 0,since derivatives are being taken with respect to variables on which the respectivefunctions do not depend.4. Prove the following identities, assuming that f is a scalar field and F is a vector field.(a) div(fF) = f div F + F · ∇f(b) curl(fF) = f curl F + (∇f) × F(c) div(∇f × ∇g) = 01Solution. We will show (a). The others follow similarly. Note thatfF = hfF1, fF2, fF3i,where F = hF1, F2, F3i. Thendiv(fF ) =∂∂x(fF1) +∂∂y(fF2) +∂∂z(fF3)= (fxF1+ fyF2+ fzF3) + (f(F1)x+ f(F2)y+ f(F3)z)= ∇f · F + f(div(F)),where we have simply used the product rule and grouped terms.5. If F = r/rp, find div F. Is there a value of p for which div F = 0?Solution. Note thatF =1rphx, y, zi.As such,∂F1∂x=∂∂xxrp=1rp−px2rp+2,and similarly∂F2∂y=1rp−py2rp+2,∂F3∂z=1rp−pz2rp+2.Putting it together,div(F ) =3rp−prp+2(x2+ y2+ z2) =3 − prp,so the divergence is zero if p = 3. Note that in these calculations we used thatrx=xr, ry=yr, rz=zr,which we calculated in a previous homework.6. Find a parametric representation for the part of the plane z = x + 3 that lies insidethe cylinder x2+ y2= 1.Solution. This is just a graph in cylindrical coordinates. We have z = r cos θ + 3, soa good parametrization isr(r, θ) = hr cos θ, r sin θ, r cos θ + 3i.Here 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π.27. Find an equation of the tangent plane to the parametric surfacer(u, v) = huv, u sin v, v cos ui,at the point u = 0, v = π.Solution. We simply follow the formula. Note thatru(u, v) = hv, sin v, −v sin ui, rv(u, v) = hu, u cos v, cos ui.And soru(0, π) = hπ, 0, 0i, rv(0, π) = h0, 0, 1i.Then a suitable normal vector isn = ru× rv= h0, −π, 0i.The point on the surface at (u, v) = (0, π) isr(0, π) = h0, 0, πi,so using the equation of a plane n · (r − r0) = 0, the equation ish0, −π, 0i · (hx, y, zi − h0, 0, πi) = 0,or−π(y − 0) = 0,ory =