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# Stokes example part 3

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

## Want to join the conversation?

• At , 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 , 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 , 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 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.
• 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?
• shouldn't you have divide the cross product with its magnitude to get the unit vector ?
• i wonder do we have to divide r/2^0.5 for the rj+rk cause it needs to be unit vector?
• 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 , 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).
• 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?
(1 vote)
• 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 ?
(1 vote)
• 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.
(1 vote)
• why isn't the normal vector a unit normal vector?
(1 vote)
• 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
(1 vote)