Math 21a Surface Area Spring, 20091In this problem, we’ll find the surface area of a sphere of radius a. We think we know thatthe answer should be 4πa2, but now we’ve defined the area of a parameterized surface to beA =ZZR|ru× rv| dA, (∗)so we should be able to make sure.(a) Use spherical coordinates to write down a parameterization of the sphere of radius a.Recall that we can use spherical coordinates to parameterize the sphere of radius a asr(φ, θ) = ha sin(φ) cos(θ), a sin(φ) sin(θ), a cos(φ)i.What values of φ and θ are needed to parameterize the entire sphere? (This will tell youthe region R over which we will integrate.)(b) Now find rφand rθusing the parameterization from part (a).(c) Compute |rφ× rθ|. You should get a2sin(φ).(d) Find the surface area of the sphere using the formula (∗).2One particular parameterization that we might take is for the graph of a function z = f (x, y).We can then replace (u, v) with (x, y), so we get the parameterizationr(x, y) = hx, y, f(x, y)i.(a) Use the above expression for r(x, y) to compute rx, ry, and |rx× ry|.(b) Use your answer to part (a) and equation (∗) to find that the surface area of the graphof the function f(x, y) isA =ZZRq1 + f2x+ f2ydA =ZZRq1 + |∇f|2dA. (∗∗)(c) Use equation (∗∗) to find the surface area of the paraboloid z = x2+ y2that lies overthe disk x2+ y2≤ 9 of radius 3.3Another particularly straightforward set of examples is surfaces of revolution. If we revolvethe graph a function f(x) (a ≤ x ≤ b) around the x-axis, we get a surface that may beparameterized byr(x, θ) = hx, f (x) cos(θ), f (x) sin(θ)i.(a) Compute rxand rθ.(b) Show that |rx× rθ| = f(x)q1 + (f0(x))2.(c) Use your answer to part (b) to deduce that the area of this surface of revolution is givenbyA = 2πZabf(x)q1 + (f0(x))2dx. (†)4Use equation (†) to compute the area of the frustrum of a coneobtained by revolving the line y = x (between x = 1 and x = 2)around the x-axis. (Note: this problem is really easy using theabove equation.).................................................................................................................................................................................................................................................................................−2−112...................................................................................................................................................................................................................................................................................................................................................................................................................................................................xy................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................5Let’s re-do Problem 1, now thinking of the sphere as a surface of revolution. Compute thesurface area of a sphere of radius a by revolving the curve y =√a2− x2(−a ≤ x ≤ a) aroundthe
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