# 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, $\iint_R (v\cos u)^2v^2\sqrt{2}vdA=\iint_R\sqrt{2}\cos^2uv^5dA=\sqrt{2}\int_0^{2\pi}\cos^2udu\int_1^3v^5dv=\frac{364}{3}\sqrt{2}\pi$.