At 7:52, how come he integrates with respect to dr d(theta) and not r dr d(theta)?

Integrating with respect to r dr d(theta) is only required when changing the variables of an integral from x and y terms to polar coordinates. In this case, however, the expression for the normal vector is already in terms of r and theta so the conversion factor of r was not required. Consider this as well though. We know that any plane with the form Ax+By+Cz+D=0 has a normal vector of Ai+Bj+Ck. Thus the plane y+z=2 (or 0x+1y+1z-2=0) will have a normal vector of 0i+1j+1k. Since the surface in question lies in that plane, it will also have a normal of 1j+1k. This vector is in terms of x y and z though, so when changing the integral to polar coordinates, r dr d(theta) is required. Therefore the double integral is of curlF(dot)(1j+1k)r dr d(theta) which is equivalent to curlF(dot)(rj+rk)dr d(theta).

At 0:30, why is ds equal to Sr x S theta dr d theta? Thank you!

Here is my intuition after analyzing the problem a little, but it is by no means a rigorous proof. S is the position vector that defines the slanted plane with respect to the origin. S can be evaluated at any (r, theta) pair to obtain a point on that plane. By picking (r, theta) as defined by the boundaries 0<r<1 and 0<theta<2*pi will limit the points to within the slanted circular area in question. Now, if you take the derivative of S with respect r, you will obtain a vector that is tangential to the slope. This is because if you evaluate S at two points where the second r is slightly larger than the first, both of the points will be on the slanted plane. In other words, the difference between the two position vectors will be parallel to the slanted plane, in the general direction of r. Same occurs with theta. Thus, at any point on S, if you take the derivative in two orthogonal directions, you will obtain a vector normal to the slanted plane by the RHR. If you integrate using the boundaries you will obtain the area of the slanted circle.

At 0:59, why is the equation not rdθdr like we learned in polar coordinates?

The integral areas are not the same. An infinitesimal area on the x-y plane (or to say the z=0 plane) can be represented in cartesian coordinate as dxdy, as well as in polar coordinate as r·drdθ. We've learned the geometric intuition behind this transformation in https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/double-integrals-a/a/double-integrals-in-polar-coordinates In the Jacobian's point of view, if we parameterize the coordinates in the x-y plane in 3D space as [x=rcosθ, y=rsinθ, z=0], or to say *V*(r,θ)=[rcosθ, rsinθ, 0], the Jacobian can be calculated as the length of the cross product *V* r x *V*θ= ``` i j k cosθ sinθ 0 -rsinθ rcosθ 0 = [0, 0, r],```thus the length is r, and it is multiplied in the integral as r·drdθ, which is consistant with the result from the geometric intuition. However in this video, we are parameterize an infinitesimal area not on the z=0 plane, but the intersection plane y+z=2, therefore it's not suitable just to multiply the r in the integral, instead, we have to evaluate the corresponding Jacobian from the cross product based on the parameterization of *S*(r,θ)=[rcosθ, rsinθ, 2-rsinθ]. And the result of the cross product is *S* r x *S*θ= ``` i j k cosθ sinθ -sinθ -rsinθ rcosθ -rcosθ = [0, r, r],```thus the length is √2 r. Then, why don't we multiply it into the integral? Because it will be cancelled out from the denominator of n̂ as we've learned in https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/flux-in-3d-articles/a/unit-normal-vector-of-a-surface Finally, only the numerator of n̂ remains, and it is the cross product of *S* r x *S*θ. Therefore the whole double integral becomes *F* · (*S* r x *S*θ) dθdr. Hope it will help.

shouldn't you have divide the cross product with its magnitude to get the unit vector ?

You're partially correct. Check this out:https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/3d-flux/v/vector-representation-of-a-surface-integral

should n-hat not be a unit vector? in this it essentially came out to (0, j-hat, k-hat) but this is great than 1 magnitude. Why is this not divided by the square rout of 2?

Because the length will be cancelled out from the denominator of n̂ as we've learned in https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/flux-in-3d-articles/a/unit-normal-vector-of-a-surface Finally, only the numerator of n̂ remains, and it is the cross product of S r x Sθ. Therefore the whole double integral becomes F · (S r x Sθ) dθdr.

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Stokes example part 3

Name: Stokes example part 3
Uploaded: 2012-06-29T01:00:41Z
Description: Converting the surface integral to a double integral

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Converting the surface integral to a double integral. Created by Sal Khan.

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ObianujuU
Posted 11 years ago. Direct link to ObianujuU's post “At 7:52, how come he inte...”
At
7:52
, how come he integrates with respect to dr d(theta) and not r dr d(theta)?
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(13 votes)
Answer
- Matt
  Posted 11 years ago. Direct link to Matt's post “Integrating with respect ...”
  Integrating with respect to r dr d(theta) is only required when changing the variables of an integral from x and y terms to polar coordinates. In this case, however, the expression for the normal vector is already in terms of r and theta so the conversion factor of r was not required.
  
  Consider this as well though. We know that any plane with the form Ax+By+Cz+D=0 has a normal vector of Ai+Bj+Ck. Thus the plane y+z=2 (or 0x+1y+1z-2=0) will have a normal vector of 0i+1j+1k. Since the surface in question lies in that plane, it will also have a normal of 1j+1k. This vector is in terms of x y and z though, so when changing the integral to polar coordinates, r dr d(theta) is required. Therefore the double integral is of curlF(dot)(1j+1k)r dr d(theta) which is equivalent to curlF(dot)(rj+rk)dr d(theta).
  Comment on Matt's post “Integrating with respect ...”
  (24 votes)
Caresse Zhu
Posted 7 years ago. Direct link to Caresse Zhu's post “At 0:30, why is ds equal ...”
At
0:30
, why is ds equal to Sr x S theta dr d theta? Thank you!
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(8 votes)
Answer
- 21duncanlo
  Posted 3 years ago. Direct link to 21duncanlo's post “Here is my intuition afte...”
  Here is my intuition after analyzing the problem a little, but it is by no means a rigorous proof.
  
  S is the position vector that defines the slanted plane with respect to the origin. S can be evaluated at any (r, theta) pair to obtain a point on that plane. By picking (r, theta) as defined by the boundaries 0<r<1 and 0<theta<2*pi will limit the points to within the slanted circular area in question.
  
  Now, if you take the derivative of S with respect r, you will obtain a vector that is tangential to the slope. This is because if you evaluate S at two points where the second r is slightly larger than the first, both of the points will be on the slanted plane. In other words, the difference between the two position vectors will be parallel to the slanted plane, in the general direction of r. Same occurs with theta. Thus, at any point on S, if you take the derivative in two orthogonal directions, you will obtain a vector normal to the slanted plane by the RHR. If you integrate using the boundaries you will obtain the area of the slanted circle.
  Button navigates to signup page
  (4 votes)
Yixuan Liu
Posted 4 years ago. Direct link to Yixuan Liu's post “At 0:59, why is the equat...”
At
0:59
, why is the equation not rdθdr like we learned in polar coordinates?
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(3 votes)
Answer
- hbliu104
  Posted 3 years ago. Direct link to hbliu104's post “The integral areas are no...”
  The integral areas are not the same.
  
  An infinitesimal area on the x-y plane (or to say the z=0 plane) can be represented in cartesian coordinate as dxdy, as well as in polar coordinate as r·drdθ.
  We've learned the geometric intuition behind this transformation in https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/double-integrals-a/a/double-integrals-in-polar-coordinates
  In the Jacobian's point of view, if we parameterize the coordinates in the x-y plane in 3D space as [x=rcosθ, y=rsinθ, z=0], or to say V(r,θ)=[rcosθ, rsinθ, 0], the Jacobian can be calculated as the length of the cross product
  V r x Vθ=
  i j k cosθ sinθ 0 -rsinθ rcosθ 0 = [0, 0, r],
  thus the length is r, and it is multiplied in the integral as r·drdθ, which is consistant with the result from the geometric intuition.
  
  However in this video, we are parameterize an infinitesimal area not on the z=0 plane, but the intersection plane y+z=2, therefore it's not suitable just to multiply the r in the integral, instead, we have to evaluate the corresponding Jacobian from the cross product based on the parameterization of S(r,θ)=[rcosθ, rsinθ, 2-rsinθ].
  And the result of the cross product is
  S r x Sθ=
  i j k cosθ sinθ -sinθ -rsinθ rcosθ -rcosθ = [0, r, r],
  thus the length is √2 r.
  Then, why don't we multiply it into the integral? Because it will be cancelled out from the denominator of n̂ as we've learned in https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/flux-in-3d-articles/a/unit-normal-vector-of-a-surface
  
  Finally, only the numerator of n̂ remains, and it is the cross product of S r x Sθ. Therefore the whole double integral becomes F · (S r x Sθ) dθdr. Hope it will help.
  Comment on hbliu104's post “The integral areas are no...”
  (6 votes)
Joshua Burke
Posted 12 years ago. Direct link to Joshua Burke's post “Why is the "cross product...”
Why is the "cross product" be thought of as two hands with fingers pointing in perpendicular directions with the thumb facing up? Is the "thumb" vector thought of in respect to the other two vectors, or one in particular?
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(3 votes)
Answer
diamantidisno3
Posted 5 years ago. Direct link to diamantidisno3's post “shouldn't you have divide...”
shouldn't you have divide the cross product with its magnitude to get the unit vector ?
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(3 votes)
Answer
- Amit Ferman
  Posted 5 years ago. Direct link to Amit Ferman's post “You're partially correct....”
  You're partially correct. Check this out:https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/3d-flux/v/vector-representation-of-a-surface-integral
  Comment on Amit Ferman's post “You're partially correct....”
  (3 votes)
Pui Tak Lam
Posted 11 years ago. Direct link to Pui Tak Lam's post “i wonder do we have to di...”
i wonder do we have to divide r/2^0.5 for the rj+rk cause it needs to be unit vector?
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(4 votes)
Answer
- Guido
  Posted 10 years ago. Direct link to Guido's post “I guess that if you use n...”
  I guess that if you use n times dS (scalar), you do need to divide it, but since you're using n times dS (vector), you do not need to do it. Go to
  4:56
  , it says: n.dS (scalar) = dS (vector) = the cross product we're using (hopefully when you see it on the vid you'll see what I'm talking about).
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  (0 votes)
Kyle Brisky
Posted 6 years ago. Direct link to Kyle Brisky's post “should n-hat not be a uni...”
should n-hat not be a unit vector? in this it essentially came out to (0, j-hat, k-hat) but this is great than 1 magnitude. Why is this not divided by the square rout of 2?
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(1 vote)
Answer
- hbliu104
  Posted 3 years ago. Direct link to hbliu104's post “Because the length will b...”
  Because the length will be cancelled out from the denominator of n̂ as we've learned in https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/flux-in-3d-articles/a/unit-normal-vector-of-a-surface
  Finally, only the numerator of n̂ remains, and it is the cross product of S r x Sθ. Therefore the whole double integral becomes F · (S r x Sθ) dθdr.
  Button navigates to signup page
  (2 votes)
Mohammed Wasay Mudassir
Posted 7 years ago. Direct link to Mohammed Wasay Mudassir's post “I guess ds is the differe...”
I guess ds is the differential area element of the projected circle on the plane and from the plane equation the normal vector to the plane is (0,1,1) if we think in that fashion then in the polar coordinates rdrd(theta) is the differential area in the xy plane but our interest is the differential area on the surface of the projected circle on the plane which would be different but the end expression which sal came with is rdrd(theta) how do we justify ?
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(1 vote)
Answer
- Amit Ferman
  Posted 5 years ago. Direct link to Amit Ferman's post “He didn't quite end up wi...”
  He didn't quite end up with this as the cross product was still present before the drd(theta). You are correct, though, in that the differential area rdrd(theta) represents a differential are upon the input space and that the differential area he showed using the cross product is the one we should be interested in.
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  (1 vote)
rushaidamahbub1
Posted a year ago. Direct link to rushaidamahbub1's post “why isn't the normal vect...”
why isn't the normal vector a unit normal vector?
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(1 vote)
Answer
sahith3353
Posted 6 years ago. Direct link to sahith3353's post “After dot product of (cur...”
After dot product of (curl f) in strokes theorem
If i get sign on 'z' is negative and given normal is positive 'z' then should i negative the dot produc
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(1 vote)
Answer

Video transcript

We're now ready to get into the meat of evaluating this surface integral. And we really need to re-express it in terms of a double integral in the domain of the parameters. And the first thing I'm going to do is rewrite this part right over here using our parameters. And we already know that n, our normal vector times our surface differential, can also be written as kind of a vector version of our surface differential that points in the same direction as our normal vector. And this is going to be the same thing. And we need to make sure that we get the order on the cross product right as the partial derivative-- and I'm going to confirm this in a second. The partial derivative of the parametrization with respect to one of the parameters crossed with the partial derivative of the parametrization with respect to the other parameter. And then that whole thing-- I'm not going to take the absolute value because I need a vector right over here-- times differentials of the parameters-- d theta dr. And we can swap these two things around depending on what will make our eventual double integral easier. But we can't swap these two things around because this would actually change the direction of the vector. So we need to make sure that this is popping us out in the right direction. So let's think about the direction that the partial with respect to r will take us. So as r increases, we're going to be moving radially outward from the center of our surface. Let me do this in a different color. As r increases, we'll be moving radially outwards. So this quantity will be a vector that looks something like that. And then as theta increases, we'll be going roughly in that direction. And so if we take the cross product of those two things-- and we could use the right-hand rule. Take your right hand, point your index finger in the direction of that yellow vector. Let me make it clear-- this is the orange vector right over here. Take your index finger in the direction of that yellow vector. So that's my index finger, my shakily drawn yellow index finger. Put your middle finger in the direction of the orange vector. So my middle, you bend it and put it in the direction of the orange vector. And we don't care what the other two fingers do. And then, your thumb will be in the direction of the cross product. So your thumb will point outward like that. That's my best attempt at drawing it. Which is exactly the direction we needed to point it. We need it to point upward here in order to be oriented properly with the direction that we are actually traversing the path. So this is actually the right order. If when we did this we got the thumb pointing into it or below the plane, then we would actually have to swap these orders. So with that out of the way, let's actually evaluate this cross product. So the cross product of the partial of our parametrization with respect to r crossed with the partial of our parametrization with respect to theta. I like to set up the matrix to take the cross product. So we'll put i, j, and k components, < like that. And then, first, I will write the partial with respect to r. So the i-component, if you take the derivative of this with respect to r, is just going to be cosine theta. The derivative of this with respect to r is just sine theta. And the derivative of this with respect to are r is just going to be a negative sine theta. And then we're going to cross it with this, so the derivative of this with respect to theta is going to be negative r sine of theta. The derivative of our j-component with respect to theta will be r cosine theta. And the derivative or our k-component-- or our z-component with respect to theta-- is going to be-- let's see-- negative r cosine theta. Is that right? Yeah. Derivative of sine theta's cosine theta. Yeah. So it's negative r cosine theta. And now we just evaluate this determinant over here. Our i-component is going to be-- so ignore that row and this column. And we get sine times negative r cosine theta. I'll just do this in a new color. So we're going to get negative r-- that wasn't a new color. I'll do it in purple. We will get negative r cosine theta sine theta minus-- well, this is going to be a negative number. So when you subtract a negative number, it's going to be plus. So it's going to be plus r cosine theta sine theta. It's always nice when things cancel out like this. This plus this is just 0. Negative of it plus the positive of it. So that all canceled out to 0. We don't have an i-component. Now, let's go to the j-component. And remember, we need to do our little checkerboard pattern. So it's going to be minus j. And it's going to be-- ignore this column, that row-- cosine theta times negative r. Cosine theta is negative r cosine squared theta. I just multiplied those two. And then, from that I'm going to subtract this times that. And so this times that, the negatives cancel out. We get r sine squared theta. So this is minus-- let me make sure. Yeah, I'm going to subtract the product of these two. The product is positive. It's going to be r sine squared theta. This is always a hard part. You can make a lot of careless mistakes here. And this looks like we might be able to simplify this in a second, but I'll wait. Actually, I'll just distribute this negative sign just for fun. So if we distribute the negative sign, these all become positive. Helps simplify things a little bit. And now let's worry about the k-component. The k-component I will be doing in purple-- well, I'll do it in blue. The k-component. Ignore this row, ignore this column. So plus k times cosine theta times r cosine theta is r cosine squared of theta. And then from that, I'm going to subtract negative r sine theta times sine theta. So that's going to be negative r sine squared theta, but I'm subtracting it. So it's going to be plus r sine squared theta. And this looks like we're going to simplify it as well. And so this piece right over here, we can factor out-- let me just rewrite it. This can be rewritten as r times cosine squared theta plus sine squared theta. Basic trig identity. That just evaluates to 1. So this is just r times j. And this over here simplifies for the exact same reason. This is r times cosine squared theta plus sine squared theta. This also is just 1. So this just simplifies to r times k. This whole cross product-- all of this business right over here-- simplified to, quite luckily, is equal to r times our j unit vector plus r times our k unit vector. And so now we can write our surface integral. Our original surface integral we can write as the double integral. Or we might want to change the order in which we integrate, but we'll give ourselves that option a little bit later. Double integral. Now, it's going to be over our parameter domain or the r theta domain. So it's the double integral of-- we still have the curl of f. And we're going to have to evaluate the curl of f. So I'll just write "curl of our vector field f" dotted with this business, rj plus rk. And then we have our two parameters. And we might want to switch the order. So maybe we could write d theta dr. And then if we do it in this order, theta goes from 0 to 2 pi, and r is going from 0 to 1. But if we swap these two, then obviously, we're going to have to swap these two as well. I'll leave you there. In the next video, we will evaluate the curl of f. And maybe in that video, if we have time, we will get to the finish line.

Multivariable calculus

Course: Multivariable calculus > Unit 5

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Video transcript