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17Calculus - Surface Integrals

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There are several equations for surface integrals and which one you use depends on what form your equations are in. This is a similar situation that you encountered with line integrals.

There are two main groups of equations, one for surface integrals of scalar-valued functions and a second group for surface integrals of vector fields (often called flux integrals). The following table places them side-by-side so that you can easily see the difference.

scalar-valued function

\(f(x,y,z)\)

vector field

\(\vec{F}(x,y,z)=M(x,y,z)\hat{i} + N(x,y,z)\hat{j} + P(x,y,z)\hat{k}\)

What Are Surface Integrals?

Before we get started with the details of surface integrals and how to evaluate them, let's watch a couple of great videos that will gently introduce you to surface integrals of scalar-valued functions and give you some examples. This is one of our favorite instructors and we think these videos are worth taking the time to watch.

Dr Chris Tisdell - Surface Integrals and Scalar-Valued Functions (1) [41mins-38secs]

video by Dr Chris Tisdell

In this second video from about the 15min-20sec mark to the end of the video, he discusses a way to simplify the integration in a special case. We suggest that you do not watch that part of the video until you have worked some practice problems and are comfortable with the basics of setting up the integrals. However, check with your instructor to see if they will allow you to use this technique.

Dr Chris Tisdell - Surface Integrals and Scalar-Valued Functions (2) [30mins-47secs]

video by Dr Chris Tisdell

Surface Integrals of Scalar-Valued Functions

The following equations are used when you are given a scalar-valued function over which you need to evaluate a surface integral. The form of the function is \(f(x,y,z)\). In general, your surface is parameterized as \(\vec{r}(u,v)=x(u,v)\hat{i} + y(u,v)\hat{j} + z(u,v)\hat{k}\). So to evaluate the integral of \(f(x,y,z)\) over the surface \(\vec{r}\), we use the equation

\( \iint\limits_S {f(x,y) ~ dS} = \iint\limits_R {f(x(u,v),y(u,v),z(u,v)) ~ \| \vec{r}_u \times \vec{r}_v \| ~ dA} \)

\(\vec{r}(u,v)\) is the parametric surface

R is the region in the uv-plane

\(\vec{r}_u\) and \(\vec{r}_v\) are the partial derivatives of \(\vec{r}\)

In the special case where we have the surface described as \(z=g(x,y)\), we can parameterize the surface as \(\vec{r}=x\hat{i}+y\hat{j}+g(x,y)\hat{k}\). This gives \( \| \vec{r}_x \times \vec{r}_y \| = \sqrt{1+[g_x]^2+[g_y]^2} \) and the surface integral can then be written as \( \iint\limits_S {f(x,y) ~ dS} = \iint\limits_R {f(x,y,z) ~ \sqrt{1+[g_x]^2+[g_y]^2} ~ dA} \) and R is the region in the xy-plane.

Okay, let's try some practice problems evaluating surface integrals of scalar functions.

Practice - Surface Integrals, Scalar-Valued Functions

Basic

Evaluate \( \iint_S { x^2 y z ~ dS } \) where S is the part of the plane \( z = 1 + 2x + 3y \) that lies above the rectangle \( 0 \leq x \leq 3, 0 \leq y \leq 2 \).

Problem Statement

Evaluate \( \iint_S { x^2 y z ~ dS } \) where S is the part of the plane \( z = 1 + 2x + 3y \) that lies above the rectangle \( 0 \leq x \leq 3, 0 \leq y \leq 2 \).

Final Answer

\( 171 \sqrt{14} \)

Problem Statement

Evaluate \( \iint_S { x^2 y z ~ dS } \) where S is the part of the plane \( z = 1 + 2x + 3y \) that lies above the rectangle \( 0 \leq x \leq 3, 0 \leq y \leq 2 \).

Solution

1881 video

video by PatrickJMT

Final Answer

\( 171 \sqrt{14} \)

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Evaluate \( \iint_S{ xy ~ dS } \) using a parametric surface where S is \( x^2 + y^2 = 4, 0 \leq z \leq 8 \) in the first octant.

Problem Statement

Evaluate \( \iint_S{ xy ~ dS } \) using a parametric surface where S is \( x^2 + y^2 = 4, 0 \leq z \leq 8 \) in the first octant.

Final Answer

32

Problem Statement

Evaluate \( \iint_S{ xy ~ dS } \) using a parametric surface where S is \( x^2 + y^2 = 4, 0 \leq z \leq 8 \) in the first octant.

Solution

1882 video

video by MIP4U

Final Answer

32

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Evaluate \( \iint_S { x^2 + y^2 } \) using a parametric surface where S is the hemisphere \( x^2 + y^2 + z^2 = 1 \) above the xy-plane.

Problem Statement

Evaluate \( \iint_S { x^2 + y^2 } \) using a parametric surface where S is the hemisphere \( x^2 + y^2 + z^2 = 1 \) above the xy-plane.

Final Answer

\(4\pi/3\)

Problem Statement

Evaluate \( \iint_S { x^2 + y^2 } \) using a parametric surface where S is the hemisphere \( x^2 + y^2 + z^2 = 1 \) above the xy-plane.

Solution

1883 video

video by MIP4U

Final Answer

\(4\pi/3\)

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Integrate \( f(x,y,z) = xy \) over the surface \( z = 4 - 2x - 2y \) in the first octant.

Problem Statement

Integrate \( f(x,y,z) = xy \) over the surface \( z = 4 - 2x - 2y \) in the first octant.

Final Answer

2

Problem Statement

Integrate \( f(x,y,z) = xy \) over the surface \( z = 4 - 2x - 2y \) in the first octant.

Solution

1995 video

video by MIP4U

Final Answer

2

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Intermediate

Compute the surface integral of \(\displaystyle{ f(x,y,z) = \frac{2z^2}{x^2+y^2+z^2} }\) over the cap of the sphere \( x^2 + y^2 + z^2 = 9 \), \(z \geq 2\).

Problem Statement

Compute the surface integral of \(\displaystyle{ f(x,y,z) = \frac{2z^2}{x^2+y^2+z^2} }\) over the cap of the sphere \( x^2 + y^2 + z^2 = 9 \), \(z \geq 2\).

Final Answer

\( 76 \pi/9 \)

Problem Statement

Compute the surface integral of \(\displaystyle{ f(x,y,z) = \frac{2z^2}{x^2+y^2+z^2} }\) over the cap of the sphere \( x^2 + y^2 + z^2 = 9 \), \(z \geq 2\).

Solution

Note - At about the 13min-50sec mark, he writes \(76\pi/3\) as the answer. At the end of the video, he corrects his answer to \(76\pi/9\).

1992 video

video by Dr Chris Tisdell

Final Answer

\( 76 \pi/9 \)

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A roof is given by the graph of \( g(x,y) = 25 + 0.5x + 0.5y \) over \( 0 \leq x \leq 40 \), \( 0 \leq y \leq 20 \). If the density of the roof is given by \( f(x,y,z) = 150 - 2z \), determine the mass of the roof.

Problem Statement

A roof is given by the graph of \( g(x,y) = 25 + 0.5x + 0.5y \) over \( 0 \leq x \leq 40 \), \( 0 \leq y \leq 20 \). If the density of the roof is given by \( f(x,y,z) = 150 - 2z \), determine the mass of the roof.

Final Answer

\( 281994\sqrt{6} \)

Problem Statement

A roof is given by the graph of \( g(x,y) = 25 + 0.5x + 0.5y \) over \( 0 \leq x \leq 40 \), \( 0 \leq y \leq 20 \). If the density of the roof is given by \( f(x,y,z) = 150 - 2z \), determine the mass of the roof.

Solution

1994 video

video by MIP4U

Final Answer

\( 281994\sqrt{6} \)

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Surface Orientation

Before we discuss surface integrals over vector fields, we need to discuss surface orientation. Surface orientation is important because we need to know which direction the vector field is pointing, outside or inside, in order to determine the flux through the surface.

The two vectors that calculated above, \(\vec{r}_u\) and \(\vec{r}_v\) and tangent vectors to the surface. Using the cross product, we can calculate two possible normal vectors, \(\vec{N}_1 = \vec{r}_u \times \vec{r}_v\) and \(\vec{N}_2 = \vec{r}_v \times \vec{r}_u\). One vector points inward, the other points outward. We will use the first one, i.e. \(\vec{N}_1 = \vec{r}_u \times \vec{r}_v\) and divide by the length to get the unit vector \(\displaystyle{ \vec{N} = \frac{\vec{r}_u \times \vec{r}_v}{ \| \vec{r}_u \times \vec{r}_v \|} }\) which is called the upward unit normal.
[Important Note: This may not always be the outward pointing normal but for our discussions we will work with surfaces where the outward pointing normal is this upward pointing normal. For your application, you will need to double check that you have the outward pointing normal.]

Surface Integrals of Vector Fields

In this video, Dr Chris Tisdell continues his discussion of surface integrals and talks about vector fields. Again, this is a great video to watch.

Dr Chris Tisdell - Surface integrals + vector fields [25mins]

video by Dr Chris Tisdell

Surface integrals over vector fields are often called flux integrals since we will often be calculating the flux through a closed surface. The flux of a vector field \(\vec{F}(x,y,z)=M(x,y,z)\hat{i}+N(x,y,z)\hat{j}+P(x,y,z)\hat{k}\) through a surface S with a unit normal vector \(\vec{N}\) is \( \iint\limits_S { \vec{F} \cdot \vec{N} ~ dS} \).
[ Note: In the above description, there are two N's. One is a function \(N(x,y,z)\) which is the j-component of the vector function. The other is \(\vec{N}\), a unit normal vector. They are distinct and unrelated and should not be confused. ]

When the surface is given in terms of \(z=g(x,y)\), we can calculate the unit normal vector as \(\displaystyle{ \frac{\nabla G}{\| \nabla G \|} }\). Since \(dS = \| \nabla G \| ~ dA \) where \(G(x,y,z)=z-g(x,y)\), the surface integral becomes \( \iint\limits_R { \vec{F} \cdot \nabla G ~ dA } \) where R is the projection of S in the xy-plane.

Surface Integrals - Meaning and Applications

The meaning of the surface integral depends on what the function \(f(x,y,z)\) or \(\vec{F}(x,y,z)\) represents. Here is a video clip giving some applications.

Evans Lawrence - Lecture 31 - Parametric Surfaces, Surface Integrals [29mins-3secs]

video by Evans Lawrence

In this final video, he gives more explanation of surface integrals and a couple of examples. He has a unique way of thinking about surface integrals and, as he says at the first of this video, surface integration is not easy. So it will help you to watch this video clip as well before going on to trying some on your own.

Evans Lawrence - Lecture 32 - More on Parametric Surfaces, Surface Integrals [23mins-15secs]

video by Evans Lawrence

Okay, you are now ready for some practice problems calculating surface integrals using vector functions.
Then you will be ready for the three dimensional versions of Green's Theorem, Stokes' Theorem and the Divergence Theorem.

Practice - Surface Integrals, Vector Functions

Basic

Compute the flux of \( \vec{F} = \langle x,y,z \rangle \) across the surface \( z = 4 - x^2 - y^2 \), \( z \geq 0 \) oriented up.

Problem Statement

Compute the flux of \( \vec{F} = \langle x,y,z \rangle \) across the surface \( z = 4 - x^2 - y^2 \), \( z \geq 0 \) oriented up.

Final Answer

\( 24\pi \)

Problem Statement

Compute the flux of \( \vec{F} = \langle x,y,z \rangle \) across the surface \( z = 4 - x^2 - y^2 \), \( z \geq 0 \) oriented up.

Solution

1993 video

video by Evans Lawrence

Final Answer

\( 24\pi \)

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Determine the flux of \( \vec{F} = \langle 0,-1,-2 \rangle \) across the surface \( z = 6 - x - y \) in the first octant. Use a downward orientation.

Problem Statement

Determine the flux of \( \vec{F} = \langle 0,-1,-2 \rangle \) across the surface \( z = 6 - x - y \) in the first octant. Use a downward orientation.

Final Answer

54

Problem Statement

Determine the flux of \( \vec{F} = \langle 0,-1,-2 \rangle \) across the surface \( z = 6 - x - y \) in the first octant. Use a downward orientation.

Solution

1996 video

video by MIP4U

Final Answer

54

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Determine the surface area of the cylinder given by \( \vec{r} = \langle 3\cos(u), 3\sin(u), v\rangle \) for \( 0 \leq u \leq 2\pi \), \( 0 \leq v \leq 4 \).

Problem Statement

Determine the surface area of the cylinder given by \( \vec{r} = \langle 3\cos(u), 3\sin(u), v\rangle \) for \( 0 \leq u \leq 2\pi \), \( 0 \leq v \leq 4 \).

Final Answer

\( 24\pi \)

Problem Statement

Determine the surface area of the cylinder given by \( \vec{r} = \langle 3\cos(u), 3\sin(u), v\rangle \) for \( 0 \leq u \leq 2\pi \), \( 0 \leq v \leq 4 \).

Solution

1997 video

video by MIP4U

Final Answer

\( 24\pi \)

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Determine the surface area of the sphere given by \( \vec{r} = \langle 2\sin(u)\cos(v), 2\sin(u)\sin(v), 2\cos(u)\rangle \) for \( 0 \leq u \leq \pi \), \( 0 \leq v \leq 2\pi \).

Problem Statement

Determine the surface area of the sphere given by \( \vec{r} = \langle 2\sin(u)\cos(v), 2\sin(u)\sin(v), 2\cos(u)\rangle \) for \( 0 \leq u \leq \pi \), \( 0 \leq v \leq 2\pi \).

Final Answer

\( 16\pi \)

Problem Statement

Determine the surface area of the sphere given by \( \vec{r} = \langle 2\sin(u)\cos(v), 2\sin(u)\sin(v), 2\cos(u)\rangle \) for \( 0 \leq u \leq \pi \), \( 0 \leq v \leq 2\pi \).

Solution

1998 video

video by MIP4U

Final Answer

\( 16\pi \)

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Intermediate

Calculate the flux of \( \vec{F} = z\hat{i} + yz\hat{j} + 2x\hat{k} \) across the upper hemisphere of the unit sphere oriented with outward-pointing normals.

Problem Statement

Calculate the flux of \( \vec{F} = z\hat{i} + yz\hat{j} + 2x\hat{k} \) across the upper hemisphere of the unit sphere oriented with outward-pointing normals.

Final Answer

\( \pi/4 \)

Problem Statement

Calculate the flux of \( \vec{F} = z\hat{i} + yz\hat{j} + 2x\hat{k} \) across the upper hemisphere of the unit sphere oriented with outward-pointing normals.

Solution

Equation

Since the flux \(F\) is defined parametrically, we will use the equation \( \iint_{S}{ \vec{F} \cdot \vec{N} ~ dS } = \iint_{D}{ \vec{F} \cdot ( \vec{r}_u \times \vec{r}_v ) ~ dA } \) where \(\vec{N}\) is the unit normal vector.

Surface

The equation of a sphere is \(x^2 + y^2 + z^2 = R^2\) where \(R\) is the radius of the sphere. In our case, we have a unit sphere, so \(R=1\). It will be easier to convert to spherical coordinates and work with them, rather than rectangular coordinates.
In general, a sphere is \( \vec{r}(\theta, \phi) = \langle R\cos\theta\sin\phi, R\sin\theta\sin\phi, R\cos\phi \rangle \)
where \(0 \leq \theta \lt 2\pi\) and \(0 \leq \phi \leq \pi\).
Since we just need the upper hemisphere of a unit sphere, our equations are \( \vec{r}(\theta, \phi) = \langle \cos\theta\sin\phi, \sin\theta\sin\phi, \cos\phi \rangle \) where \(0 \leq \theta \lt 2\pi\) and \(0 \leq \phi \leq \pi/2\).

To go along with the notation in the tutorial, we will use the variables \(u\) and \(v\) instead of \(\theta\) and \(\phi\), where \(u=\theta\) and \(v=\phi\). This means the equation for the surface is \( \vec{r}(u, v) = \langle R\cos u \sin v, R\sin u \sin v, R\cos v \rangle \) where \(0 \leq u \lt 2\pi\) and \(0 \leq v \leq \pi/2\).

Calculating The Partial Derivatives

We need to determine \( \vec{r}_u \times \vec{r}_v \) where \( \vec{r}_u \) is the partial derivative of \( \vec{r} \) with respect to \(u\). Similarly for the second term.

\( \vec{r}_u = \cos u \cos v \hat{i} + \) \( \cos v \sin u \hat{j} - \) \( \sin v \hat{k} \)

\( \vec{r}_v = -\sin v \sin u \hat{i} + \) \( \sin v \sin u \hat{j} + 0 \hat{k} \)

Calculating The Cross Product

The zero in the \(\hat{k}\) term of the second partial derivative helps to make the cross product not as complicated.

\( \vec{r}_u \times \vec{r}_v = \) \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \cos u \cos v & \cos v \sin u & -\sin v \\ -\sin u \sin v & \sin u \sin v & 0 \end{vmatrix} \) \( = \sin u \sin^2 v \hat{i} + \) \( \sin u \sin^2 v \hat{j} + \) \( ( \cos u \cos v \sin u \sin v + \cos v \sin u \sin^2 v ) \hat{k} \)

Calculating The Dot Product

Okay, so now we need to take the result from last line above and take the dot of that with the flux vector, i.e. we need to calculate \( \vec{F} \cdot ( \vec{r}_u \times \vec{r}_v ) \)

The result, after a bit of simplifying is \( \cos v \sin^2 v \sin u + \cos v \sin^3 v \sin^2 u + 2\cos v \sin^2 v \cos^2 u \sin u + 2\cos v \sin ^2 v \cos u \sin^2 u \)

Integrating

Whew! Now we need to integrate that last equation twice, once with respect to \(u\) and again with resepect to \(v\). That seems like a lot of work. However, since the limits of integration in each case are all constants, we can easily choose the order of integration. In all four cases, we chose to integrate with respect to \(u\) first. This causes the integration of the first, third and fourth integrals to all go to zero without even needing to integrate in the other direction. The only integral that results in a non-zero value is the second one. We will go through that one and let you work the other three.

The inside integral is \(\displaystyle{ \int_0^{2\pi}{ \sin^2 u ~ du } }\). Substituting \(\displaystyle{ \sin^2 u = \frac{1-\cos(2u)}{2} }\) and integrating, gives us \( \pi \).

The outside integral is then \(\displaystyle{ \int_0^{\pi/2}{ \pi \cos v \sin^3 v ~dv } }\)

Pull a \(\sin v\) out of the cubed term, substitute \(\sin^2 u = 1- \cos^2 u\) and using integration by substitution yields \(\pi/4\) for the integral.

Final Answer

\( \pi/4 \)

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