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Surface Integral

Surfaces and surface areas are commonly used in integration. Surface Integrals are to be related to surface area much the same way as the line integral is related to arc length. Surface integral evaluates a function defined over a surface. The surface in a surface integral is the level surface for the integrand function.

The surface integral is said to be the generalization integration over surface for multiple integrals. This integral can be imagined as an analog of double integral for the line integral. We may integrate for the scalar fields or vector fields of a given surface. The surface integrals are found in a number of areas such as physics, generally in classical electromagnetism. In this article, we are going to learn about the concept of surface integral and various examples based on it.

Related Calculators
Calculate Integral Calculate Definite Integrals
Calculate Double Integral Calculate Surface Area
 

Evaluation of Surface Integral

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Given below is an examples for the evaluation of Surface Integral.

Solved Example

Question: Evaluate the integral $\int \int _{s}x^{2}yz ds$, where S is the part of the plane z = 1 + 2x + 3y and D is the region 0 $\leqslant$ x $\leqslant$ 3 , 0 $\leqslant$ y $\leqslant$ 2.
Solution:
$\int \int _{s}x^{2}yz ds = \int \int _{D}x^{2}yz\sqrt{(\frac{\partial z}{\partial x})^{2}+(\frac{\partial z}{\partial y})^{2}+1}dA$

The partial derivatives of the surface are $\frac{\partial z}{\partial x}$$=2$ and $\frac{\partial z}{\partial y}$$=3$

$\int \int _{s}x^{2}yz ds = \int \int _{D}x^{2}yz$$\sqrt{(\frac{\partial z}{\partial x})^{2}+(\frac{\partial z}{\partial y})^{2}+1}dA$

$=$$\int_{0}^{3}\int_{0}^{2}x^{2}y(1+2x+3y)\sqrt{4+9+1}dydx$

$=\sqrt{14}$$\int_{0}^{3}\int_{0}^{2}$$(x^{2}y+2x^{3}y+3x^{2}y^{2})dydx$

$=\sqrt{14}$$\int_{0}^{3}$$[\frac{1}{2}x^{2}y^{2}+x^{3}y^{2}+x^{2}y^{3}]_{y=0}^{y=2}dx$
$=\sqrt{14}$$\int_{0}^{3}$$(10x^{2}+4x^{3})dx$
$=\sqrt{14}$$[\frac{10}{3}x^{3}+x^{4}]_{0}^{3}$$ = 171\sqrt{14}$

If needed the double integral can also be evaluated using polar coordinates.

Surface Integral Sphere

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A three dimension body which is bounded by any surface and whose every points are equidistant from a fixed is called sphere. The fixed point is known as the center of the sphere and the distance which is same is called the radius of the sphere. The general equation of sphere is (x - a)$^2$ + (y - b)$^2$ + (z - c)$^2$ = r$^2$
Where, (a,b,c) are the coordinates of the center of sphere and r is the radius of the sphere.

We can find out the total surface area of the sphere(4$\pi$ r$^2$) by the use of integral(surface integral).
We can use the formula for surface integral as:

$\int \int _{S}f(x,y,z)dS = \int \int _{D}f(x,y,g(x,y))$ $\sqrt{(\frac{\partial z}{\partial x})^{2}+(\frac{\partial z}{\partial y})^{2}+1}$$dA$

For this we assume that
S is a surface z = g(x,y) and S is the projected on XY plane and g has continuous first partial derivatives.

Surface Integral Spherical Coordinates

We know that in spherical coordinate

$x = \rho \cos \theta \sin \phi , y = \rho \sin \theta \sin \phi , z = \rho \cos \phi$

$\rho \rightarrow 0$ to $2 \pi $ and $\phi \rightarrow 0 to \pi$

Here, $\rho$ is the radius in spherical coordinate system. Let $\psi (\phi, \theta)$ and
$d \sigma = \parallel \frac{\partial \psi }{\partial \theta}\times \frac{\partial \psi }{\partial \phi}\parallel d\theta \phi$

Then, surface integral becomes

$\int _{\sum }F(x,y,z)d\sigma$ = $\int _{D}$$F\left ( \rho (\theta ,\phi ) \cos \theta \sin \phi, ( \rho (\theta ,\phi ) \sin \theta \sin \phi, ( \rho (\theta ,\phi ) \cos \phi \right )$$\rho \sqrt{\left ( \rho ^{2} + \frac{\partial \rho ^{2}}{\partial \theta } \sin ^{2}\theta + \frac{\partial \rho ^{2}}{\partial \phi }\right )}d\theta d\phi$


If we have 
$x^2$ + $y^2$ +$z^2$ $a^2$ then $r^2$ = $a^2$ 


So, r = $\rho(\theta,\phi)$ = a

d$\sigma$ = a$^2$$\sin \theta d\theta d\phi$

$\int _{\sum }d\sigma $ = $\int \int _{D}a^{2} \sin \theta d\theta d\phi$

= $\int_{0}^{2\pi}\int_{0}^{\pi}a^{2} \sin \theta d\theta d\phi$

= $4 \pi a^{2}$


Example

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Given below are some of the examples on Surface Integral.

Solved Examples

Question 1: Compute the Surface Integral $\int _{S}\int x^{2}dS$ where S is the unit sphere $x^{2}+y^{2}+z^{2}=1$
Solution:
We can use the spherical coordinates for the parametrization. As r = 1, we have
x = $\cos (\theta) \sin (\phi), y = \sin (\theta) \sin(\theta) \ and\ $ z = $\cos (\phi).\ 0 \leqslant \phi \leqslant 2 \pi.$

The vector equation of the unit sphere is thus

$r(\phi,\theta) = \sin (\phi) cos(\theta)\ i + \sin(\phi) sin(\theta) j + \cos (\phi)k$

$r_{\phi }Xr_{\theta } = \begin{vmatrix}
i & j & k\\
\cos \phi \cos \theta & \cos \phi \sin \theta &-\sin \phi \\
-\sin \phi \sin \theta &\sin \phi \cos \theta & 0
\end{vmatrix}$


=$\sin^2\phi \cos\theta\ i + \sin^2\phi \sin\theta\ j + sin\theta\ \cos\phi k$ 

Hence, we can calculate $|r_{\phi }Xr_{\theta }|$ as $|r_{\phi }Xr_{\theta }| = \sin \phi $

Using the formula for the vector parametrization

$\int _{S}\int x^{2} ds = \int \int _{D}(\sin \phi \cos \theta )^{2}|r_{\phi }Xr_{\theta }|dA$

$=$$\int_{0}^{2\pi }\int_{0}^{\pi }\sin ^{2}\phi \cos ^{2}\theta \sin \phi d\phi d\theta $

$=$$\int_{0}^{2\pi }\cos ^{2}\theta d\theta \int_{0}^{\pi }\sin ^{3}\phi d\phi $

$=$$\int_{0}^{2\pi }\frac{1}{2}(1 + \cos 2\theta )d\theta \int_{0}^{\pi }(1 - \cos ^{2}\phi )\sin \phi d\phi $

$=$$\frac{1}{2}[\theta +\frac{1}{2}\sin 2\theta ]_{0}^{2\pi }[-\cos \phi +\frac{1}{3}\cos ^{3}\phi ]_{0}^{\pi }$

$=$$(\pi )(\frac{4}{3})=\frac{4\pi }{3}$

Question 2: Compute the Flux of the vector field F(x,y,z) = over the positively oriented surface of z = $x^2$ + $y^2$ under the plane z =4.
Solution:
Here, we use the surface integral for vector field formula as

$\int \int _{S}F.dS=\int \int _{D}(-P$$\frac{\partial g}{\partial x}$ $- Q$$\frac{\partial g}{\partial y}$$ + R)dA$

Given F = xi + yj (P = x, Q = y and R = 0) and $\frac{\partial g}{\partial x}$$ = 2x$ , $\frac{\partial g}{\partial y}$$ = 2y$.

$\int \int _{S}F.dS = \int \int _{D}(-P$$\frac{\partial g}{\partial x}$$ - Q$$\frac{\partial g}{\partial y}$$+ R)dA$

$= \int \int _{D}(-x(2x) - y(2y) + 0)dA$

$= \int \int _{D}(-2x^{2} - 2y^{2})dA$

The Shadow region of the surface is the circle $x^2$ + $y^2$ = 4

Using the parametrization x = 2 cos θ and y = 2 sin θ, the double integral can be evaluated as follows:

$=$ $\int_{0}^{2\pi }\int_{0}^{2}$$(-2r^{2})rdrd\theta = -16\pi $

Surface Integral Cylinder

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To find the surface integral of cylinder, first we can find out the area of top, area of bottom and area of sides.

Surface Integral Cylindrical Coordinate:

This is the extension of the polar coordinate system from two dimension to three dimension system. We know that r and $\theta$ are the polar coordinates. But, in three dimension system, we can add one coordinate say z on it. Cylindrical coordinate system is the three dimension coordinate system. So, we have r, $\theta$ and z are the coordinate in this system.

From Cartesian to Cylindrical system
x = r cos $\theta$, y = r sin$\theta $ and z = z
If we can use these relations, then we will convert any formula from Cartesian to cylindrical system. From Cartesian to cylindrical system
  • r is the radial distance from z-axis to point  P on the curve.
  • $\theta$ is the angle between direction on the line from the origin to P on the plane and the selected plane.
  • z is the height from the plane to the point P.
With the help of these information, we can convert the formula for surface integral from cartesian to cylindrical system.

Solved Example

Question: Calculate $\int \int _{S}(2x^{3}\textbf{i} +2y^{3}\textbf{j}).dS$, where S is the cylinder bounded by two surfaces, one is circle $x^2$ + $y^2$ = 4 and on top and bottom by the plane z = 0 and z = 6 oriented by outward vectors.
Solution:
From above statement, it is clear that the S is sum of the three smooth pieces:
  1. The cylindrical surface say S1 which is given by $x^2$ + $y^2$ = 4 and having normal unit vector n1 as $\frac{2x\textbf{i}+2y\textbf{j}}{\sqrt{4x^{2}+ 4y^{2}}} = \frac{x\textbf{i}+y\textbf{j}}{\sqrt{x^{2}+ y^{2}}} = \frac{x\textbf{i}+y\textbf{j}}{2}$
  2. The bottom surface S2 which is a portion of z = 0 and taking unit normal vector n2 = - k.
  3. The top surface S3 which is a portion of z = 6 and taking unit normal vector n2 = k.
Surface Integral Cylinder
Now,
$\int \int _{S}(2x^{3}\textbf{i}+2y^{3}\textbf{j}). dS = \int \int _{S_{1}}(2x^{3}\textbf{i}+2y^{3}\textbf{j}). dS + \int \int _{S_{2}}(2x^{3}\textbf{i}+2y^{3}\textbf{j}). dS + \int \int _{S_{3}}(2x^{3}\textbf{i}+2y^{3}\textbf{j}). dS$

= $\int \int _{s_{1}}(2x^{3}\textbf{i}+2y^{3}\textbf{j}).$$(\frac{x\textbf{i}+y\textbf{j}}{2})$$dS + \int \int _{s_{2}}(2x^{3}\textbf{i}+2y^{3}\textbf{j}).(-\textbf{k})dS + \int \int _{s_{3}}(2x^{3}\textbf{i}+2y^{3}\textbf{j}).(\textbf{k})dS$

= $\int \int _{S_{1}}(x^{4}+y^{4}). dS + 0 + 0$

If we take x = 2 cos u , y = 2 sin u and z = t where 0 $\leq$ u $\leq$ 2$\pi$ and 0 $\leq$ t $\leq$ 6. Then, the integral becomes

$\int \int _{S}(2x^{3}\textbf{i}+2y^{3}\textbf{j}). dS$ = $\int_{0}^{6}\int_{0}^{2\pi}(16 \cos ^{4}u + 16 \sin ^{4}u). 2 du dt$

= $32\int_{0}^{6}\int_{0}^{2\pi}(16 \cos ^{4}u + 16 \sin ^{4}u)du dt$

$32 \int_{0}^{6}\int_{0}^{2\pi}\left [ \left ( \frac{1+ \cos 2u}{2} \right )^{2} + \left ( \frac{1 - \cos 2u}{2} \right )^{2}\right ]du dt$


= $8\int_{0}^{6}\int_{0}^{2\pi}(2 + 2 \cos ^{2}2u)du dt$

= $8\int_{0}^{6}\int_{0}^{2\pi}\left [ 2 + \left ( 1 + \cos 4u \right ) \right ]du dt$

= $8\int_{0}^{6}\int_{0}^{2\pi}\left [ 3 + \cos 4u \right ]du dt$

= $8\int_{0}^{6}\left [ 3u + sin 4u \right ]_{0}^{2\pi}$

= $8\int_{0}^{6}6\pi dt $

= 48$\pi \left ( t \right )_{0}^{6}$

= $48 \pi \times 6$

= $288 \pi$

Surface Integral of a Cone

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To find the surface integral of a cone, we have to find the area of its base which is a circle and area of curved surface.

Solved Example

Question: If S is the total area of the cone $\sqrt{x^{2}+y^{2}}\leq z\leq 3$ then calculate $\int \int _{S}2z^{2}.dS $
Solution:
Let $S_1$ be the base and $S_2$ be the curved surface of the given cone. Then, we divide the total area of the cone in to two parts
I = $I_1$ + $I_2$ 
= $\int \int _{S_{1}} 2z^{2}dS_{1} + \int \int _{S_{2}}2z^{2}dS_{2}$

First we calculate I1. For this, the equation z = 3 describes the base of the given cone. Hence,
$I_1$ = $\int \int _{S_{1}} 2z^{2}dS_{1}$

= 2$\int \int _{S_{1}} 3^{2}dS_{1}$

= 18 $\int \int _{S_{1}} dS_{1}$

where, $\int \int _{S_{1}} dS_{1}$ is the area of the base which is $\pi$ $r^2$ = $3^2$ $\pi$ = 9 $\pi$.
Then, $I_1$ = 18 $\int \int _{S_{1}} dS_{1}$ = 18 (9$\pi$) = 162$\pi$

Again, $I_2$ = $\int \int _{S_{2}}2z^{2}dS_{2}$

= $\int \int _{D}f\left [ x,y,z(x,y) \right ]\sqrt{1 + \frac{\partial z}{\partial x}^{2}+\frac{\partial z}{\partial y}^{2}} $ dx dy

Here, $\frac{\partial z}{\partial x}$$ =$ $\frac{x}{\sqrt{x^{2}+y^{2}}}$ and $\frac{\partial z}{\partial y}$$ = $$\frac{y}{\sqrt{x^{2}+y^{2}}}$

Therefore,
$\sqrt{1 + \frac{\partial z}{\partial x}^{2}+\frac{\partial z}{\partial y}^{2}}$ = $\sqrt{2}$

Here, the domain D is defined by x2 + y2 $\leq$ 9 (since z = 3 on the base of the given cone). So,

$I_2$ = $\int \int _{S_{2}}2z^{2}dS_{2}= \sqrt{2}\int \int _{D}(x^{2}+y^{2})dx dy$

Take x = r cos $\theta$ , y = r sin $\theta$ then dx dy = r dr d$\theta$

Therefore, the integral becomes

$I_2$ = $\sqrt{2}\int \int _{D}(x^{2}+y^{2})dx dy$

= $\sqrt{2}$$\int_{0}^{2\pi}\int_{0}^{3}$$r^{2}. r dr d\theta$

= $\sqrt{2}$$\int_{0}^{2\pi}$$d\theta$ $\int_{0}^{3}r^{3}$$ dr $

= $\sqrt{2}(2\pi)\left( \frac{r^{4}}{4} \right )_{0}^{3}$

= $\sqrt{2} \pi$ . $\frac{81}{2}$

Then, the total surface area of the cone is I = $I_1$ + $I_2$ 

= 162 $\pi$ + $\sqrt{2} \pi$ . $\frac{81}{2}$

Surface Integral of a Cube

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To find the surface area of a cube, we can find out the area of one side of the cube and multiply that with six(6). Here, we can use the formula $\int \int _{S}F.dS = \int \int \int _{C}\triangledown  .F dV$
In above formula, S is the surface of the cube.

Solved Example

Question: Find the surface area S of a cube whose vertices are (0,0,0), (1,0,0),(0,1,0) and (0,0,1) and whose normal is outward pointing and where F(x,y,z) = xy i + $y^2$ j + xz k
Solution:
Here, we apply the divergence theorem for finding the surface area of the cube, which is:
$\int \int _{S}F.dS = \int \int \int _{C}\triangledown .F dV$

First, we calculate $\triangledown .\textbf{F} = \triangledown .(xy \textbf{i} + y^{2}\textbf{j} + xz \textbf{k})$
= (x + 3y)
So, we get
$\int \int \int _{C}\triangledown . \textbf{F} dV = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(x+3y)dx dy dz$ = 2

For surface area S, we marked six faces as $S_1$, $S_2$, $S_3$, $S_4$, $S_5$ and $S_6$ and for these faces, we have $S_1$ is x = 1, $S_2$ is x = 0 , $S_3$ is y = 1, $S_4$ is y = 0, $S_5$ is z = 1 and $S_6$ is z = 0.
Now, for $S_1$ 

$\int \int _{S_{1}}\textbf{F}.dS = \int_{0}^{1}\int_{0}^{1}(y\textbf{i}+ y^{2}\textbf{j}+ z\textbf{k}).(\textbf{i}dy dz)$ = $\frac{1}{2}$ .................................................(i)

For $S_2$

$\int \int _{S_{2}}\textbf{F}.dS = \int_{0}^{1}\int_{0}^{1}(0\textbf{i}+ y^{2}\textbf{j}+ 0\textbf{k}).(\textbf{-i}dy dz)$ = 0 .....................................................(ii)

For $S_3$ 

$\int \int _{S_{3}}\textbf{F}.dS = \int_{0}^{1}\int_{0}^{1}(x\textbf{i}+ \textbf{j}+ xz\textbf{k}).(\textbf{j}dy dz)$ = 1 .......................................................(iii)

For $S_4$

$\int \int _{S_{4}}\textbf{F}.dS = \int_{0}^{1}\int_{0}^{1}(0\textbf{i}+ 0\textbf{j}+ xz\textbf{k}).(\textbf{-j}dy dz)$ = 0 ........................................................(iv)

For $S_5$

$\int \int _{S_{5}}\textbf{F}.dS = \int_{0}^{1}\int_{0}^{1}(xy\textbf{i}+ y^{2}\textbf{j}+ x\textbf{k}).(\textbf{k}dy dz)$ = $\frac{1}{2}$ .......................................................(v)

For $S_6$

$\int \int _{S_{6}}\textbf{F}.dS = \int_{0}^{1}\int_{0}^{1}(xy\textbf{i}+ y^{2}\textbf{j}+ 0\textbf{k}).(\textbf{-k}dy dz)$ = 0 ....................................................(vi)
Adding all these equations, we get

$\int \int _{S}F.dS$ = $\int \int _{S_{1}}\textbf{F}.dS + \int \int _{S_{2}}\textbf{F}.dS + \int \int _{S_{3}}\textbf{F}.dS + \int \int _{S_{4}}\textbf{F}.dS + \int \int _{S_{5}}\textbf{F}.dS + \int \int _{S_{5}}\textbf{F}.dS$

= $\frac{1}{2}$ + 0 + 1 + 0 + $\frac{1}{2}$ + 0

= 2.

Practice Problems

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Here are given some problems of surface integral for practice.

Problem 1 : If S be a surface defined by $x^{2} + y^{2}$ = 9, 0 $\leq$ z $\leq$ 4. Find the value of integral $\int \int_{S}(x + y + z) dA$

Problem 2 : Find $\int_{S} r dS$, where $r^{2} = x^{2} + y^{2} + z^{2}$. It is given that S be the surface of unit cube, with 0 $\leq$ x $\leq$ 1, 0 $\leq$ y $\leq$ 1, 0 $\leq$ z $\leq$ 1.

Problem 3 :  Evaluate the integral $\int \int_{S} (x + y) dA$. Given that S is the region in the interior of a triangle whose vertices are (x, y) = (0, 0), (0, 1) and (2, 0).
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