# Recitation 15

Section 13.6 Problem 20: Find a parametric representation for the surface. The part of the sphere x^2 +y^2 +z^2 =16 that lies between the planes z=-2 and z=2.

Solution 1: Let z=v. Then x^2+y^2=16-v^2. Therefore x=\sqrt{16-v^2}\cos u, y=\sqrt{16-v^2}\sin u with 0\leq u \leq 2\pi, -2\leq v\leq 2.

Solution 2: If we take the spherical coordinate, x=4\sin\phi\cos\theta, y=4\sin\phi\sin\theta, z=4\cos\phi with 0\leq\theta\leq 2\pi. However, in order to have -2\leq z\leq 2, i.e., -2\leq 4\cos\phi \leq 2, \frac{\pi}{3} \leq \phi \leq \frac{2\pi}{3}.

Section 13.6 Problem 30: Find an equation of the tangent plane to the given parametric surface at the specified point. x = u^2 + 1, y = v^3 + 1, z = u + v; (5, 2, 3)

Solution: When u=2, v=1, r(u,v)=(u^2+1,v^3+1,u+v)=(5,2,3). The normal vector of the tangent plane is given by r_u(2,1)\times r_v(2,1). Since r_u=(2u,0,1), r_v=(0,3v^2,1), r_u(2,1)\times r_v(2,1)=(4,0,1)\times (0,3,1)=(-3,-4,12). Therefore, the equation of the tangent plane is -3(x-5)-4(y-2)+12(z-3)=0.

Section 13.6 Problem 37: Find the area of the surface. The surface z=\frac{2}{3}(x^{3/2} +y^{3/2}), 0\leq x\leq 1, 0\leq y\leq 1.

Solution: The area of the surface is given by \iint_R\sqrt{F_x^2+F_y^2+1}dA, where R is given by 0\leq x\leq 1, 0\leq y\leq 1 and F=\frac{2}{3}(x^{3/2} +y^{3/2}). Warning: this formula is only applicable when the surface is given by z=F(x,y). Therefore the area of the surface is \iint_R\sqrt{x+y+1}dA=\int_0^1\int_0^1\sqrt{x+y+1}dydx=\frac{4}{9}(1-4\sqrt{2}+3\sqrt{3}).

Section 13.6 Problem 38: Find the area of the surface. The part of the surface z=1+3x+2y^2 that lies above the triangle with vertices (0, 0), (0, 1), and (2, 1).

Comment: Denote the triangle with vertices (0, 0), (0, 1), and (2, 1) by T. The area is \iint_T\sqrt{F_x^2+F_y^2+1}dA, where F=1+3x+2y^2. Therefore the area of the surface is given by the following double integral \iint_T\sqrt{3^2+(4y)^2+1}dA. Then turn this double integral to an intreated integral.

Section 13.6 Problem 42: Find the area of the surface. The helicoid (or spiral ramp) with vector equation r(u , v) = u \cos v i + u \sin v j + v k , 0 \leq u \leq 1, 0\leq v\leq \pi

Solution: We know r_u = (\cos v, \sin v, 0), r_v = (-u\sin v, u\cos v, 1) and r_u\times r_v = (\sin v, \cos v, u), |r_u\times r_v|=\sqrt{u^2+1}. The area of the suface is given by \iint_R \sqrt{1+u^2}dA, where R is given by 0 \leq u \leq 1, 0\leq v\leq \pi. By Fubini’s Theorem, this double integral is equal to \int_0^1\sqrt{1+u^2}du\int_0^\pi dv. Using integral by parts, we obtain \begin{aligned}\int_0^1\sqrt{1+u^2}du & = \left.\sqrt{1+u^2}u\right|_0^1 - \int_0^1ud\sqrt{1+u^2}\\ & = \sqrt{2} - \int_0^1\frac{u^2}{\sqrt{1+u^2}}du\\ & = \sqrt{2} - \int_0^1\frac{u^2+1-1}{\sqrt{1+u^2}}du \\ & = \sqrt{2} - \int_0^1\frac{u^2+1}{\sqrt{1+u^2}}du + \int_0^1\frac{1}{\sqrt{1+u^2}}du.\end{aligned}

Notice that the second term in the last line is exactly \int_0^1\sqrt{1+u^2}du. Therefore 2\int_0^1\sqrt{1+u^2}du = \sqrt{2}+\int_0^1\frac{1}{\sqrt{1+u^2}}du = \sqrt{2}+\sinh^{-1}(1). Hence, the area is \frac{\pi}{2}(\sqrt{2}+\sinh^{-1}(1)).

Section 13.7 Problem 13: Evaluate the surface integral. \iint_S x^2z^2dS, S is the part of the cone z^2 =x^2 +y^2 that lies between the planes z=1 and z=3.

Solution: Parametrize the surface by x=v\cos u, y=v\sin u, z=v, where 0\leq u\leq 2\pi, 1\leq v\leq 3. We have r_u=(-v\sin u, v\cos u, 0), r_v=(\cos u, \sin u, 1). Therefore r_u\times r_v=(v\cos u, v\sin u, -v) and |r_u\times r_v|=\sqrt{2}v. Now we can turn the surface integral into a double integral, \iint_R (v\cos u)^2v^2\sqrt{2}vdA=\iint_R\sqrt{2}\cos^2uv^5dA=\sqrt{2}\int_0^{2\pi}\cos^2udu\int_1^3v^5dv=\frac{364}{3}\sqrt{2}\pi.