Surface Integrals

Curves and surfaces

Curves in the 2D plane

• Explicit form: $y=f(x)$

• Implicit form: $F(x,y)=0$

• Parametric vector form: $\vec r(t)=f(t)\vec i+g(t)\vec j$, $a\leq t\leq b$

Surfaces in the 3D space

• Explicit form: $z = f(x, y)$

• Implicit form: $F(x, y, z) = 0$.

• Parametric vector form: ?

Parametrizations of surfaces

Suppose that $$\vec r(u,v)=f(u,v)\vec i+g(u,v)\vec j+h(u,v)\vec k$$ is a continuous vector function defined on a region $R$ in the $uv$-plane and one-to-one on the interior of $R$.

• The range of $\vec r$ is the surface $S$ defined or traced by $\vec r$.

• The equation, together with the domain $R$, constitutes a parametrization of the surface.

• $u$ and $v$ are parameters, and $R$ is the parameter domain.

Example: Find a parametrization of the cone using $(r,\theta)$ $$z=\sqrt{x^2+y^2},\quad 0\leq z\leq 1.$$

Solution: $$\vec r(r,\theta)=(r\cos\theta)\vec i+(r\sin \theta)\vec j+r\vec k.\quad 0 \leq r \leq 1,\quad 0 \leq \theta \leq 2\pi.$$

Example: Find a parametrization of the sphere using $(\phi,\theta)$. $$x^2 + y^2 + z^2 = a^2$$

Solution: $$\vec r(\phi,\theta)=(a \sin \phi \cos \theta)\vec i + (a \sin \phi \sin \theta)\vec j + (a \cos \phi) \vec k.\quad 0 \leq \phi \leq \pi,\quad 0 \leq \theta \leq 2\pi.$$

Example: Find a parametrization of the cylinder: $$x^2+(y-3)^2=9,\qquad 0\leq z\leq 5.$$

Solution: $$\vec r(\theta,z)=(3\sin{2\theta})\vec i + (6\sin^2\theta)\vec j + (z) \vec k.\quad 0 \leq \theta \leq \pi,\quad 0 \leq z \leq 5.$$

Smooth surface

A curved surface $S$ is expressed as $$\vec r(u, v) = f(u, v)\vec i + g(u, v)\vec j + h(u, v)\vec k,~a \leq u\leq b,~c \leq v \leq d.$$ The partial derivatives are $$\begin{align*} \vec r_u={\partial \vec r\over\partial u}={\partial f\over\partial u}\vec i+{\partial g\over\partial u}\vec j+{\partial h\over\partial u}\vec k\\ \vec r_v={\partial \vec r\over\partial v}={\partial f\over\partial v}\vec i+{\partial g\over\partial v}\vec j+{\partial h\over\partial v}\vec k \end{align*}$$

Definition A parametrized surface $\vec r(u, v) =f(u, v)\vec i + g(u, v)\vec j + h(u, v)\vec k$ is smooth if $\vec r_u$ and $\vec r_v$ are continuous and $\vec r_u\times \vec r_v$ is never zero on the interior of the parameter domain (make sure that a tangent plane exists).

Surface area

Definition The area of the smooth surface $$\vec r(u, v) = f(u, v)\vec i + g(u, v)\vec j + h(u, v)\vec k,~a \leq u\leq b,~c \leq v \leq d$$ is $$A=\iint_R|\vec r_u\times \vec r_v|dA=\int_c^d\int_a^b|\vec r_u\times\vec r_v|dudv=\int_c^d\int_a^b{\color{red}d\sigma}.$$

Example: Find the surface area of the cone

Solution: We have $$\vec r(r,\theta)=(r\cos\theta)\vec i+(r\sin\theta)\vec j+r\vec k,\quad 0\leq r\leq 1,~0\leq\theta\leq 2\pi.$$

Thus $$|\vec r_r \times \vec r_{\theta}|=\sqrt{2}r.$$

Example: Find the surface area of a sphere of radius $a$.

Solution: We have $$\vec r(\phi,\theta)=(a\sin\phi\cos\theta)\vec i+(a\sin\phi\sin\theta)\vec j+(a\cos\phi)\vec k,\quad 0\leq \phi\leq \pi,~0\leq \theta\leq 2\pi,$$ and $|\vec r_\phi\times\vec r_\theta|=a^2\sin\phi$. Thus $$Area = \int_0^{2\pi}\int_0^{\pi} a^2\sin \phi d\phi d\theta = 4\pi a^2$$

Implicit surface

Surfaces are often presented as level sets of a function, described by an equation such as $F(x,y,z)=c$ for some constant c.

Formula for the Surface Area of an Implicit Surface

The area of the surface $F(x, y, z) = c$ over a closed and bounded plane region $R$ is $$\mbox{Surface area} = \iint_R{|\nabla F|\over |\nabla F\cdot \vec p|}dA$$ where $\vec p=\vec i$, $\vec j$, or $\vec k$ is normal to $R$ and $\nabla F\cdot \vec p\neq 0$.

Example: Find the area of the surface cut from the bottom of the paraboloid $x^2 + y^2 - z = 0$ by the plane $z = 4$.

Solution:

At any point $(x,y,z)$ on the surface, we have $$\nabla F = 2x \vec i + 2y \vec j - \vec k$$ $$|\nabla F| = \sqrt{4x^2 + 4y^2 + 1}$$ $$|\nabla F \cdot \vec p| = |\nabla F \cdot \vec k| = |-1| = 1.$$ Thus surface area

Surface Integrals

Applications

• the mass of a surface

• total electrical charge on a surface

• flow of a liquid across a curved membrane

Two forms

• a scalar function over a surface

• vector fields over a surface (flux)

Surface integral of $G$ over the surface $S$ $$\iint_SG(x,y,z)d\sigma$$

• If $\vec r(u,v)=f(u,v)\vec i+g(u,v)\vec j+h(u,v)\vec k$, $(u,v)\in R$, we have $$\iint_SG(x,y,z)d\sigma=\iint_RG(f(u,v),g(u,v),h(u,v)){\color{red}|\vec r_u\times \vec r_v|}dudv.$$

• $z=f(x,y)$ $$\iint_SG(x,y,z)d\sigma=\iint_RG(x,y,f(x,y)){\color{red}\sqrt{f_x^2+f_y^2+1}}dxdy.$$

• Implicitly $F(x,y,z)=c$ $$\iint_SG(x,y,z)d\sigma=\iint_RG(x,y,z){\color{red}|\nabla F|\over |\nabla F\cdot \vec p|}dA,$$ where $\vec p$ is a unit vector normal to $R$ and $\nabla F\cdot \vec p\neq 0$.

Example: Integrate $G(x, y, z) = xyz$ over the surface of the cube cut from the first octant by the planes $x = 1$, $y = 1$, and $z = 1$

Solution: There are six sides and we compute the surface integral over the six sides one by one.

• Side A: $$\begin{align*} \int_0^1\int_0^1 xydxdy={1\over 4} \end{align*}$$

• Side B: $$\begin{align*} \int_0^1\int_0^1 yzdydz={1\over 4} \end{align*}$$

• Side C: $$\begin{align*} \int_0^1\int_0^1 xzdxdz={1\over 4} \end{align*}$$

• The surface integral over the other three sides are all 0, and the total surface integral is only 3/4.

Example: Integrate $G(x,y,z)=x^2$ over the cone $z=\sqrt{x^2+y^2}$, $0\leq z\leq 1$

Solution:

Example: Integrate $G(x,y,z)=\sqrt{1-x^2-y^2}$ over the ''football'' suface $S$ formed by rotating the curve $x=\cos z$, $y=0$, $-\pi/2\leq z\leq \pi/2$, around the $z$-axis

Solution:

Example: Evaluate $\iint_S \sqrt{x(1 + 2z)} d\sigma$ on the portion of the cylinder $z = y^2/2$ over the triangular region $R: x\geq 0,~y\geq 0,~x + y \leq 1$ in the $xy$-plane

Solution:

Mass and moment formulas for very thin shells

• Mass: $$M=\iint_S\delta d\sigma,\quad \delta=\delta(x,y,z)\text{ is the density at }(x,y,z)$$

• First moments about the coordinate planes: $$M_{yz}=\iint_Sx\delta d\sigma,\quad M_{xz}=\iint_Sy\delta d\sigma,\quad M_{xy}=\iint_Sz\delta d\sigma$$

• Coordinates of the center of mass: $$\bar x=M_{yz}/M,\quad \bar y=M_{xz}/M,\quad \bar z=M_{xy}/M$$

• Moments of inertia about axes and other straight lines $$I_x=\iint_S(y^2+z^2)\delta d\sigma,\quad I_y=\iint_S(x^2+z^2)\delta d\sigma,\quad I_z=\iint_S(x^2+y^2)\delta d\sigma$$ $$I_L=\iint_Sr^2\delta d\sigma,\quad r(x,y,z)=\text{distance from the point }(x,y,z)\text{ to line }L$$

Example: Find the center of mass of a thin hemispherical shell of radius $a$ and constant density $\delta$.

Solution:

Example: Find the center of mass of a thin shell of density $\delta=1/z^2$ cut from the cone $z=\sqrt{x^2+y^2}$ by the planes $z=1$ and $z=2$

Solution:

Vector integral on a surface

The surface is in 3D. How do we generalize the line integral to 3D?

Recall that in 2D, we have the following two integrals.

• $\int_C\vec{F}\cdot\vec{T}ds$

• $\int_C\vec{F}\cdot\vec{n}ds$

Orientation of a surface

A smooth surface $S$ is orientable (or two-sided) if it is possible to define a field of unit normal vectors $\vec n$ on $S$, which varies continuously with position. By convention, we usually choose $\vec n$ on a closed surface to point outward.

Once $\vec n$ has been chosen, we say that we have oriented the surface, and we call it with its normal field an oriented surface. At any point, the vector $\vec n$ is called the positive direction.

Surface integrals of vector fields

Definition Let $\vec F$ be a vector field in three-dimensional space with continuous components defined over a smooth surface $S$ having a chosen field of normal unit vectors $\vec n$ orienting $S$. Then the surface integral of $\vec F$ over $S$ is $$\iint_S\vec F\cdot\vec n d\sigma$$ The surface integral of $\vec F$ is also called the flux of the vector field across the oriented surface $S$.

Example: Find the flux of $\vec F=yz\vec i+x\vec j-z^2\vec k$ through the parabolic cylinder $y = x^2,~0 \leq x \leq 1,~0 \leq z \leq 4$, in the direction $\vec n$ indicated in the figure.

Solution: $$\begin{align*} \vec r(x,z)=&x\vec i+x^2\vec j+z\vec k,~0\leq x\leq1,0\leq z\leq 4\\ \vec r_x\times \vec r_z = & 2x\vec i-\vec j\\ \vec n= &{2x\vec i-\vec j\over\sqrt{4x^2+1}} \end{align*}$$

Simplification for the parametrized case

For the parametrized case $$d\sigma = |\vec r_u\times \vec r_v|dudv$$and $$\vec n={\vec r_u\times \vec r_v\over |\vec r_u\times \vec r_v|}$$

For the level surface $g(x,y,z)=c$, we have $$\vec n={\pm}{\nabla g\over |\nabla g|}$$ $$\iint_S\vec F\cdot\vec n d\sigma=\iint_R(\vec F\cdot {\pm \nabla g\over |\nabla g|}){|\nabla g|\over{|\nabla g \cdot \vec p|}}dA=\iint_R\vec F\cdot{\pm \nabla g \over |\nabla g \cdot \vec p|} dA$$

Example: Find the flux of $\vec F=yz\vec j+z^2\vec k$ outward through the surface $S$ cut from the cylinder $y^2+z^2=1$, $z\geq 0$, by the planes $x=0$ and $x=1$.

Solution: