Flux integral computationsPROBLEM 1. Set up and evaluate the flux integralZ ZSF · dSwhere F(x, y, z) = (x, y, z) and S is the surface z = 4 − x2− y2defined for x2+ y2≤ 4,with the upward normal orientation.Solution. This surface is the graph of the function f(x, y) = 4− x2− y2for x2+y2≤ 4,so we shall use the graph parametrization X(x, y) = (x, y, 4 − x2− y2).We shall use the following (very important!) general formula for the upward normalof a surface defined by an equation of the form z = f (x, y):N(x, y) = (−f1. − f2, 1)For the surface of interest here, it follows thatN(x, y) =−(−2x), −(−2y), 1= (2x, 2y, 1) .Since we also haveFX(x, y)= (x, y, 4 − x2− y2)it follows that the flux F · N is given by 2x2+ 2y2+ 4 − x2− y2= 4 + x2+ y2. Thereforethe flux integral is equal toZ ZD4 + x2+ y2dAwhere D is the disk defined by x2+ y2≤ 4. One simple way to evaluate this integral is bymeans of polar coordinates, and it follows that the flux integral must be equal toZ2π0Z20(4 + r2) r dr dθ .Evaluating this iterated integral from the inside out, we see that it is equal toZ2π0Z204r +r3dr dθ =Z2π02r2+r4420dθ =Z2π0(8 + 4) dθ = 2π · 12 = 24π .PROBLEM 2. (A modified version of Problem 15 on page 438 of the text). Let S bethe boundary of the cylinder defined by x2+ y2≤ 9 and 0 ≤ z ≤ 4, and take the preferredorientation pointing outward from S. Compute the flux integral of F(x, y, z) = (x, y, z+1)over S.Solution. This surface has three separate smooth pieces, and we need to check thateach of them is parametrized so that the normal is pointing outward.(i) Top face of the cylinder. This is given by x2+ y2≤ 9 and z = 4, so it canbe parametrized as the graph of f (x, y) = 4. Over this piece of the surface, theoutward pointing normal direction is the same as the upward pointing direction,so the given parametrization yields the outward pointing normal, and if we applythe formula from the preceding section we see that N = (0, 0, 1) on this piece ofthe surface.(ii) Lateral face of the cylinder. This is given by x2+ y2= 9 and 0 ≤ z ≤ 4,so it can be parametrized by X(θ, z) = (3 cos θ, 3 sin θ, z) for 0 ≤ θ ≤ 2π and0 ≤ z ≤ 4. If we compute the normal direction for this parametrization, weobtain N = (3 cos θ, 3 sin θ, 0). This is the outward pointing normal on the lateralface of the surface.(iii) Bottom face of the cylinder. This is given by x2+ y2≤ 9 and z = 0, so it canbe parametrized as the graph of f (x, y) = 0. Over this piece of the surface, theoutward pointing normal direction is the downward pointing direction, and asin the discussion of the top face the given parametrization defines the upwarddirection. Therefore we have to adjust the parametrization so that its normaldirection points downward. One simple way of reversing the normal direction isto replace the usual parametrization (x, y, 0) with (x, −y, 0), and we choose thisparametrization for the bottom part so that its standard normal direction pointsdownward/outward.To find the flux integral over the entire cylinder, we need to compute the flux integralsfor each of the three smooth faces and conclude by adding their values together.On the top face, we know that FX(x, y)= (x, y, 5) and N = (0, 0, 1), so that(F · N) = 5. Therefore the flux over the top face is the integral of the constant function 5over the region x2+ y2≤ 9; the value of this integral is equal to 45π (using the formulaA = πr2for the area of a solid disk of radius r).On the lateral face, we have FX(θ, z)= (3 cos θ, 3 sin θ, z) and (as noted before)N = (3 cos θ, 3 sin θ, 0), so that F · N is equal to 9. Therefore the flux over the top face isthe integral of the constant function 9 over the region 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 4; by thestandard formula for the lateral area of a cylinder, the area of the lateral surface is equalto (2π) × 3 × 4 = 24π, and therefore the value the integral is equal to (24π) × 9 = 216π.Finally, on the bottom face, we know that FX(x, y)= (x, −y, 1) andN = X1× X2= (1, 0, 0) × (0, −1, 0) = (0, 0, −1)so that (F·N) = −1. Therefore the flux over the bottom face is the integral of the constantfunction −1 over the region x2+ y2≤ 9; by the same reasoning employed for the top face,we see that the value of this integral is equal to −9π.COMPUTING THE TOTAL FLUX. Having computed the flux over each of the threepieces (with the correct preferred orientations!), we can find the total flux by adding thevalues obtained for the three pieces. Therefore the total flux is equal to the sum45π + 216π − 9π = 252π