I could only understand the parameterization of ( x = r cos 𝜽 ; y = r sin 𝜽 ; z = 0 ) in the xy-plane from (0 to 2 π ) around z-axis. Could you please explain the parameterization of the varying radius from (0 to π). How do you get ( x y z ) as ( r cos sin; r sin sin; r cos) or (r cos cos; r sin sin; r sin) ?

You can take your inspiration there : https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/modal/v/determining-a-position-vector-valued-function-for-a-parametrization-of-two-parameters Think of how works spherical coordinates, and then try to find x, y and z depending on s (angle between the radius and axis z), and t, angle between the projection of the radius over the xy plane and the x axis. First the magnitude of the radius projection over the xy plane will be : rsin(s) . At this step you could find z=rcos(s) => z done. Second step : finding x and y, you take the cos(t).(projection of the radius over the xy plane)=x wich is x=cos(t)rsin(s)and sin(t).(projection of the radius over the xy plane)=y whose the equation is finally y=sin(t)rsin(s); sorry for my english, hope it helps

can someone explain why we have taken s only from 0 to pie and t 0 to 2pie

s tells us how far up we need to go in the z-direction, and once you know your z-coordinate you trace out a circle parallel to the xy-plane on top of the sphreres surface. Since you can reach every point between -2 and 2 by letting s range from 0 to pi (remember, your z-component is 2cos(s) and cos reaches every value between -1 and 1 if you use every s between 0 and pi) any other value of s would only let you trace out a circle you already traced out For visuals on what I just said open the answer for the first concept check in step 2

Here in this Sal parameterized the sphere as: [ cos(t)cos(s) , cos(t)sin(s) , sin(t) ] Why is it so different? Does it give the same result? https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/surface-integrals-introduction/v/surface-integral-example-part-1-parameterizing-the-unit-sphere

Yes, both approaches will yield that same final result, assuming the dimensions of the sphere are consistent and the functions that are being integrated over, are identical. It all hinges on how you define your s and t angles and their corresponding datums. This in turn affects the parametrization of the system. Sal's method defines his base circle in the x-y plane in terms of angle 0 < s < 2pi rotating about the z-axis. He then defines the varying radius of that circle in terms of the angle -pi/2 < t < pi/2 rotating about an arbitrary axis in the x-y plane. Sal seems to have set the x-y plane as his datum for angle t, such that t=0 in this plane. The above method seems to define the base circle in the x-y plane as well, but in terms of angle 0 < t < 2pi rotating about the z-axis. The varying radius of that circle is then defined in terms of the angel 0 < s < pi rotating about an arbitrary axis in the x-y plane. This method however has defined the positive z-axis as a datum, such that s = 0 in the positive vertical orientation. This is my understanding of the methods in these two solutions. I hope my explanation is clear enough for the distinction to be apparent to you. Remember that each mothod will have slightly different integral boundaries due to the choice of datum. Perhaps an experienced mathematician can do a better job of verballising this xD

What are Volume Integrals? ∰ Does the aforementioned symbol have any use or connection in that concept, if so what does that symbol mean?

The circle/oval through the integral symbol doesn't formally change anything. Rather, it's a suggestion that the area being integrated over is somehow "closed." For example, a line integral over a circle would typically have a circle drawn through it because the circle is a closed curve. A double integral over the surface of a sphere might have the circle through it. A triple integral over the volume of a sphere might have the circle through it. (By the way, triple integrals are often called volume integrals when the integrand is 1.) I hope this helps you make sense of the notation.

Main content

Course: Multivariable calculus > Unit 4

Lesson 12: Surface integrals (articles)

Surface integral example

Q: What are Volume Integrals? ∰ Does the aforementioned symbol have any use or connection in that concept, if so what does that symbol mean?

The circle/oval through the integral symbol doesn't formally change anything. Rather, it's a suggestion that the area being integrated over is somehow "closed." For example, a line integral over a circle would typically have a circle drawn through it because the circle is a closed curve. A double integral over the surface of a sphere might have the circle through it. A triple integral over the volume of a sphere might have the circle through it. (By the way, triple integrals are often called volume integrals when the integrand is 1.) I hope this helps you make sense of the notation.

Google Classroom

Practice computing a surface integral over a sphere.

Background

Surface integrals

The task at hand: Surface integral on a sphere.

In the last article, I talked about what surface integrals do and how you can interpret them. Here, you can walk through the full details of an example. If you prefer videos you can also watch Sal go through a different example.

Consider the sphere of radius

2

, centered at the origin.

Your task will be to integrate the following function over the surface of this sphere:

$f (x, y, z) = (x - 1)^{2} + y^{2} + z^{2}$ ‍

Step 1: Take advantage of the sphere's symmetry

The sphere with radius

2

is, by definition, all points in three-dimensional space satisfying the following property:

$x^{2} + y^{2} + z^{2} = 2^{2}$ ‍

This expression is very similar to the function:

$f (x, y, z) = (x - 1)^{2} + y^{2} + z^{2}$ ‍

In fact, we can use this to our advantage...

Concept check: When you evaluate

f (x, y, z) = (x - 1)^{2} + y^{2} + z^{2}

on points that happen to be on the sphere with radius

2

, what simpler expression do you get?

Keep in mind,

f (x, y, z)

does not equal this simpler expression everywhere, but only on the points where

x^{2} + y^{2} + z^{2} = 4

. Since we will only integrate over points on this sphere, though, we can justifiably replace the function

f

in the integral with this value.

$\begin{array}{r} \iint_{Sphere} ((x - 1)^{2} + y^{2} + z^{2}) d Σ = \iint_{Sphere} (- 2 x + 5) d Σ \end{array}$ ‍

Of course, this is not something you can do for every surface integral, but it's a good lesson to take advantage of symmetry when you can to make these integrals easier.

Step 2: Parameterize the sphere

To relate this surface integral to a double integral on a flat plane, we need to first find a function which parameterizes the sphere.

Concept check: Which of the following functions parameterizes the sphere with radius

2

Choose 1 answer:
(Choice A)
$\begin{array}{r} \vec{v} (t, s) = [\begin{array}{c} 2 \cos (t) \sin (s) \\ 2 \sin (t) \sin (s) \\ 2 \cos (s) \end{array}] \end{array}$ ‍
in the region where $0 \leq t \leq 2 π$ ‍ and $0 \leq s \leq π$ ‍.
(Choice B)
$\begin{array}{r} \vec{v} (t, s) = [\begin{array}{c} 2 \cos (t) \cos (s) \\ 2 \sin (t) \sin (s) \\ 2 \sin (t) \cos (s) \end{array}] \end{array}$ ‍
in the region where $0 \leq t \leq 2 π$ ‍ and $0 \leq s \leq 2 π$ ‍.

The first choice is correct:
$\begin{array}{r} \vec{v} (t, s) = [\begin{array}{c} 2 \cos (t) \sin (s) \\ 2 \sin (t) \sin (s) \\ 2 \cos (s) \end{array}] \end{array}$ ‍
And you apply this to the region of the $t s$ ‍-plane where $0 \leq t \leq 2 π$ ‍ and $0 \leq s \leq π$ ‍.
There are no two ways about it, parameterizing surfaces is hard. The trick to a problem like this, where you need to recognize what surface a given function will parameterize, is to think about what happens when you freeze one variable and let the other one vary.
For example, in the expression for $\vec{v} (t, s)$ ‍ above, imagine freezing $s$ ‍ and let $t$ ‍ vary:
$\begin{array}{r} [\begin{array}{c} 2 \cos (t) \sin (s) \\ 2 \sin (t) \sin (s) \\ 2 \cos (s) \end{array}] \end{array}$ ‍
Since $x$ ‍ is proportional to $\cos (t)$ ‍, and $y$ ‍ is proportional to $\sin (t)$ ‍, letting $t$ ‍ range from $0$ ‍ to $2 π$ ‍ will draw a circle around the $z$ ‍-axis:
Khan Academy video wrapper
See video transcript
Alternatively, imagine fixing $t$ ‍ and letting $s$ ‍ vary:
$\begin{array}{r} [\begin{array}{c} 2 \cos (t) \sin (s) \\ 2 \sin (t) \sin (s) \\ 2 \cos (s) \end{array}] \end{array}$ ‍
This one also draws a circle (can you see why?)
Khan Academy video wrapper
See video transcript
Now imagine letting $s$ ‍ range from $0$ ‍ to $π$ ‍, meaning only half of the red circle shown in that last animation is drawn. Rather than thinking of $t$ ‍ as a fixed amount, picture the entire circle that $t$ ‍ draws for each specific value of $s$ ‍. These circles will sweep over the sphere from top to bottom:
Khan Academy video wrapper
See video transcript
This is why we let $t$ ‍ range from $0$ ‍ to $2 π$ ‍, but we only let $s$ ‍ range from $0$ ‍ to $π$ ‍.

Great! Now we have a formula for the parameterization

\vec{v} (t, s)

of the sphere, along with a corresponding region on the

t s

-plane. We can start expanding out surface integral like this:

$\begin{aligned} \iint_{Sphere} (- 2 x + 5) d Σ \\ = \int_{0}^{π} \int_{0}^{2 π} (- 2 \underset{x -value of parameterization}{\underset{⏟}{(2 \cos (t) \sin (s))}} + 5) \underset{We need work this out}{\underset{⏟}{| \frac{\partial \vec{v}}{\partial t} \times \frac{\partial \vec{v}}{\partial s} |}} d t d s \end{aligned}$ ‍

Step 3: Compute both partial derivatives

The main beast to wrangle with in any surface integral is this little guy:

$\begin{array}{r} | \frac{\partial \vec{v}}{\partial t} \times \frac{\partial \vec{v}}{\partial s} | \end{array}$ ‍

Concept check: To start, compute both partial derivatives of our parametric function:

$\begin{array}{r} \vec{v} (t, s) = [\begin{array}{c} 2 \cos (t) \sin (s) \\ 2 \sin (t) \sin (s) \\ 2 \cos (s) \end{array}] \end{array}$ ‍

$\frac{\partial \vec{v}}{\partial t} (t, s) =$ ‍
$\hat{i} +$ ‍
$\hat{j} +$ ‍
$\hat{k}$ ‍

$\frac{\partial \vec{v}}{\partial s} (t, s) =$ ‍
$\hat{i} +$ ‍
$\hat{j} +$ ‍
$\hat{k}$ ‍

Step 4: Compute the cross product

Compute the cross product of the two partial derivative vectors that you just found.

$\frac{\partial \vec{v}}{\partial t} \times \frac{\partial \vec{v}}{\partial s} =$ ‍
$\hat{i} +$ ‍
$\hat{j} +$ ‍
$\hat{k}$ ‍

$\begin{aligned} \frac{\partial \vec{v}}{\partial t} \times \frac{\partial \vec{v}}{\partial s} \\ = [\begin{array}{c} - 2 \sin (t) \sin (s) \\ 2 \cos (t) \sin (s) \\ 0 \end{array}] \times [\begin{array}{c} 2 \cos (t) \cos (s) \\ 2 \sin (t) \cos (s) \\ - 2 \sin (s) \end{array}] \end{aligned}$ ‍
Now apply the usual determinant trick for cross products:
$\begin{array}{r} det ([\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 2 \sin (t) \sin (s) & 2 \cos (t) \sin (s) & 0 \\ 2 \cos (t) \cos (s) & 2 \sin (t) \cos (s) & - 2 \sin (s) \end{array}]) \end{array}$ ‍
We take this one component at a time.
$\hat{i}$ ‍ component: Cross out the top row and left column, then take the determinant:
$\begin{array}{r} det ([\begin{array}{cc} 2 \cos (t) \sin (s) & 0 \\ 2 \sin (t) \cos (s) & - 2 \sin (s) \end{array}]) = - 4 \cos (t) \sin^{2} (s) \end{array}$ ‍
$\hat{j}$ ‍ component: Cross out the top row and middle column, then take the negative determinant:
$\begin{array}{r} - det ([\begin{array}{cc} - 2 \sin (t) \sin (s) & 0 \\ 2 \cos (t) \cos (s) & - 2 \sin (s) \end{array}]) = - 4 \sin (t) \sin^{2} (s) \end{array}$ ‍
$\hat{k}$ ‍ component: Lastly, and most nastily, cross out the top row and last column, then take the determinant:
$\begin{aligned} det ([\begin{array}{cc} - 2 \sin (t) \sin (s) & 2 \cos (t) \sin (s) \\ 2 \cos (t) \cos (s) & 2 \sin (t) \cos (s) \end{array}]) \\ = \underset{factor out - 4 \sin (s) \cos (s)}{\underset{⏟}{- 4 \sin^{2} (t) \sin (s) \cos (s) - 4 \cos^{2} (t) \sin (s) \cos (s)}} \\ = - 4 \sin (s) \cos (s) (\underset{1}{\underset{⏟}{\sin^{2} (t) + \cos^{2} (t)}}) \\ = - 4 \sin (s) \cos (s) \end{aligned}$ ‍

Step 5: Find the magnitude of the cross product.

Find the magnitude of the cross product that you just found.

$| \frac{\partial \vec{v}}{\partial t} \times \frac{\partial \vec{v}}{\partial s} | =$ ‍

$\begin{aligned} | \frac{\partial \vec{v}}{\partial t} \times \frac{\partial \vec{v}}{\partial s} | \\ = | [\begin{array}{c} - 4 \cos (t) \sin^{2} (s) \\ - 4 \sin (t) \sin^{2} (s) \\ - 4 \sin (s) \cos (s) \end{array}] | \\ = \sqrt{(- 4 \cos (t) \sin^{2} (s))^{2} + (- 4 \sin (t) \sin^{2} (s))^{2} + (- 4 \sin (s) \cos (s))^{2}} \\ = \sqrt{\underset{Factor out 4^{2} from the radical}{\underset{⏟}{4^{2} \cos^{2} (t) \sin^{4} (s) + 4^{2} \sin^{2} (t) \sin^{4} (s) + 4^{2} \sin^{2} (s) \cos^{2} (s)}}} \\ = 4 \sqrt{\underset{Factor out \sin^{4} (s)}{\underset{⏟}{\cos^{2} (t) \sin^{4} (s) + \sin^{2} (t) \sin^{4} (s)}} + \sin^{2} (s) \cos^{2} (s)} \\ = 4 \sqrt{\sin^{4} (s) (\underset{1}{\underset{⏟}{\cos^{2} (t) + \sin^{2} (t)}}) + \sin^{2} (s) \cos^{2} (s)} \\ = 4 \sqrt{\underset{Factor out \sin^{2} (s)}{\underset{⏟}{\sin^{4} (s) + \sin^{2} (s) \cos^{2} (s)}}} \\ = 4 \sqrt{\sin^{2} (s) (\underset{1}{\underset{⏟}{\sin^{2} (s) + \cos^{2} (s)}})} \\ = 4 \sin (s) \end{aligned}$ ‍

Notice, technically the answer should have an absolute value sign in it. However, because our parameterization only applies to the region where

0 \leq s \leq π

, the value of

\sin (s)

will always be positive anyway, so we are free to leave that out.

Step 6: Compute the integral

Taking everything we've done so far, here's what the surface integral has turned into:

$\begin{aligned} \iint_{Sphere} f (x, y, z) d Σ \\ = \iint_{Sphere} (- 2 x + 5) d Σ \leftarrow Step 1 \\ = \int_{0}^{π} \int_{0}^{2 π} (- 2 (2 \cos (t) \sin (s)) + 5) | \frac{\partial \vec{v}}{\partial t} \times \frac{\partial \vec{v}}{\partial s} | d t d s \leftarrow Step 2 \\ = \int_{0}^{π} \int_{0}^{2 π} (- 2 (2 \cos (t) \sin (s)) + 5) (4 \sin (s)) d t d s \leftarrow Steps 3, 4, 5 \\ = \int_{0}^{π} \int_{0}^{2 π} (- 16 \cos (t) \sin^{2} (s) + 20 \sin (s)) d t d s \end{aligned}$ ‍

As a the final step, compute this double integral.

$\begin{array}{r} \int_{0}^{π} \int_{0}^{2 π} (- 16 \cos (t) \sin^{2} (s) + 20 \sin (s)) d t d s = \end{array}$ ‍

First, let's break this integral into two simpler ones:
$\begin{aligned} \int_{0}^{π} \int_{0}^{2 π} (- 16 \cos (t) \sin^{2} (s) + 20 \sin (s)) d t d s \\ = - 16 \int_{0}^{π} \int_{0}^{2 π} \cos (t) \sin^{2} (s) d t d s + 20 \int_{0}^{π} \int_{0}^{2 π} \sin (s) d t d s \end{aligned}$ ‍
Now let's go through each, one at a time:
$\begin{aligned} \int_{0}^{π} \int_{0}^{2 π} \cos (t) \sin^{2} (s) d t d s \\ = \int_{0}^{π} {[\sin (t) \sin^{2} (s)]}_{t = 0}^{t = 2 π} d s \\ = \int_{0}^{π} (\sin (2 π) \sin^{2} (s) - \sin (0) \sin^{2} (s)) d s \\ = \int_{0}^{π} (0 - 0) d s \\ = 0 \end{aligned}$ ‍
Well, that makes things easier! What about the other integral:
$\begin{aligned} \int_{0}^{π} \int_{0}^{2 π} \underset{Constant with respect to t}{\underset{⏟}{\sin (s)}} d t d s \\ = \int_{0}^{π} \sin (s) (2 π) d s \\ = 2 π [- \cos (s)]_{s = 0}^{s = π} \\ = 2 π (- \cos (π) - (- \cos (0))) \\ = 2 π (2) \\ = 4 π \end{aligned}$ ‍
Therefore, our final integral simplifies as
$\begin{aligned} - 16 \underset{= 0}{\underset{⏟}{\int_{0}^{π} \int_{0}^{2 π} \cos (t) \sin^{2} (s) d t d s}} + 20 \underset{= 4 π}{\underset{⏟}{\int_{0}^{π} \int_{0}^{2 π} \sin (s) d t d s}} \\ = 80 π \end{aligned}$ ‍

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Arunabh Bhattacharya
Posted 7 years ago. Direct link to Arunabh Bhattacharya's post “I could only understand t...”
I could only understand the parameterization of ( x = r cos 𝜽 ; y = r sin 𝜽 ; z = 0 ) in the xy-plane from (0 to 2 π ) around z-axis.
Could you please explain the parameterization of the varying radius from (0 to π). How do you get ( x y z ) as ( r cos sin; r sin sin; r cos) or (r cos cos; r sin sin; r sin) ?
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(4 votes)
Answer
- guilhem.escudero
  Posted 6 years ago. Direct link to guilhem.escudero's post “You can take your inspira...”
  You can take your inspiration there : https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/modal/v/determining-a-position-vector-valued-function-for-a-parametrization-of-two-parameters
  Think of how works spherical coordinates, and then try to find x, y and z depending on s (angle between the radius and axis z), and t, angle between the projection of the radius over the xy plane and the x axis. First the magnitude of the radius projection over the xy plane will be : rsin(s) . At this step you could find z=rcos(s) => z done. Second step : finding x and y, you take the cos(t).(projection of the radius over the xy plane)=x wich is x=cos(t)rsin(s)and sin(t).(projection of the radius over the xy plane)=y whose the equation is finally y=sin(t)rsin(s); sorry for my english, hope it helps
  Comment on guilhem.escudero's post “You can take your inspira...”
  (2 votes)
pranavkarel22
Posted 5 years ago. Direct link to pranavkarel22's post “can someone explain why w...”
can someone explain why we have taken s only from 0 to pie and t 0 to 2pie
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(2 votes)
Answer
- Robert
  Posted 5 years ago. Direct link to Robert's post “s tells us how far up we ...”
  s tells us how far up we need to go in the z-direction, and once you know your z-coordinate you trace out a circle parallel to the xy-plane on top of the sphreres surface. Since you can reach every point between -2 and 2 by letting s range from 0 to pi (remember, your z-component is 2cos(s) and cos reaches every value between -1 and 1 if you use every s between 0 and pi) any other value of s would only let you trace out a circle you already traced out
  
  For visuals on what I just said open the answer for the first concept check in step 2
  Comment on Robert's post “s tells us how far up we ...”
  (4 votes)
Gavaskar
Posted 8 years ago. Direct link to Gavaskar's post “Can you please help me so...”
Can you please help me solving this surface area integral
Find the surface area of the portion of the sphere of radius 4,centre at origin that lies inside the cylinder x^2+y^2=12 and above the xy-plane.
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(2 votes)
Answer
Chris Duvivier
Posted 7 years ago. Direct link to Chris Duvivier's post “Here in this Sal paramete...”
Here in this Sal parameterized the sphere as:
[ cos(t)cos(s) , cos(t)sin(s) , sin(t) ]
Why is it so different? Does it give the same result?
https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/surface-integrals-introduction/v/surface-integral-example-part-1-parameterizing-the-unit-sphere
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(1 vote)
Answer
- Jason
  Posted 7 years ago. Direct link to Jason's post “Yes, both approaches will...”
  Yes, both approaches will yield that same final result, assuming the dimensions of the sphere are consistent and the functions that are being integrated over, are identical. It all hinges on how you define your s and t angles and their corresponding datums. This in turn affects the parametrization of the system.
  
  Sal's method defines his base circle in the x-y plane in terms of angle 0 < s < 2pi rotating about the z-axis. He then defines the varying radius of that circle in terms of the angle -pi/2 < t < pi/2 rotating about an arbitrary axis in the x-y plane. Sal seems to have set the x-y plane as his datum for angle t, such that t=0 in this plane.
  
  The above method seems to define the base circle in the x-y plane as well, but in terms of angle 0 < t < 2pi rotating about the z-axis. The varying radius of that circle is then defined in terms of the angel 0 < s < pi rotating about an arbitrary axis in the x-y plane. This method however has defined the positive z-axis as a datum, such that s = 0 in the positive vertical orientation.
  
  This is my understanding of the methods in these two solutions. I hope my explanation is clear enough for the distinction to be apparent to you. Remember that each mothod will have slightly different integral boundaries due to the choice of datum. Perhaps an experienced mathematician can do a better job of verballising this xD
  Comment on Jason's post “Yes, both approaches will...”
  (3 votes)
festavarian2
Posted 2 years ago. Direct link to festavarian2's post “Unnecessarily confusing, ...”
Unnecessarily confusing, because early on, he doesn't define what the parameters t and s represent: ie. he never states that t represents the delta theta in the x-y plane and that s is phi in the spherical coordinate system.
Why didn't he do this?
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(2 votes)
Answer
Rafae Tariq
Posted 5 years ago. Direct link to Rafae Tariq's post “What are Volume Integrals...”
What are Volume Integrals?

∰

Does the aforementioned symbol have any use or connection in that concept, if so what does that symbol mean?
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(1 vote)
Answer
- lakern
  Posted 4 years ago. Direct link to lakern's post “The circle/oval through t...”
  The circle/oval through the integral symbol doesn't formally change anything. Rather, it's a suggestion that the area being integrated over is somehow "closed." For example, a line integral over a circle would typically have a circle drawn through it because the circle is a closed curve. A double integral over the surface of a sphere might have the circle through it. A triple integral over the volume of a sphere might have the circle through it. (By the way, triple integrals are often called volume integrals when the integrand is 1.) I hope this helps you make sense of the notation.
  Button navigates to signup page
  (2 votes)
Arunabh Bhattacharya
Posted 7 years ago. Direct link to Arunabh Bhattacharya's post “I would also like to unde...”
I would also like to understand Volume Integrals.
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(1 vote)
Answer