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When calculating surface integral in scalar field, we use the following formula: surface integral formula for scalar field

Now, in the example solved below;example of surface integral

For surface S1, they have calculated $r_(theta) × r_z$ for finding the normal vector, which is clear to me. But my question is whether we can use gradient to find the same normal vector and then its modulus to be replaced in the integral because in my class, our teacher has used gradient for finding the unit normal vector in many examples in surface integrals over vector field given by the formulasurface integral over vector field

Now, if I calculate the gradient of the surface I get n= 2x i+ 2y j and |n| = 2 instead of 1 found out by $r_(theta) × r_z$ in the solved question above. As a result if I substitute 2 in place of |$r_(theta) × r_z$| the whole integral value gets multiplied by 2 which certainly gives the wrong answer. I am really confused as to why the magnitude of the normal vector differs in the two cases and how to differentiate where to use the gradient and where the cross-product for calculations. Also, if normals can be found using gradient, then why at all we have been introduced with the method of using the cross product which is quite tedious to find in some cases, to find the same?

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2 Answers 2

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If the surface $S$ is given as a level set of some function $f$, then its normal is in fact parallel to the gradient of $f$.

But $\mathbf{r}_u \times \mathbf{r}_v$ has information about magnitude in it which cannot be obtained from $\nabla f$. The fact that $\nabla f$ doesn't have this information is easy to understand: if $S$ is given by $f=c$, then it is also given by $af=ac$ for any real number $a$, and the functions $af$ have different gradients. They point in the same direction (or exactly opposite directions for negative $a$) but they have entirely different magnitudes.

In short, what the gradient can do for you (for surfaces given as level sets) is to find $\mathbf{n}$. This is OK if you can use some geometry to perform an integral $dS$, but that is usually not possible. The gradient cannot tell you how to convert $dS$ into $|J(u,v)| du dv$, however.

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  • $\begingroup$ Then, which magnitude does $r_u × r_v$ gives actually? $\endgroup$ Commented Aug 24, 2020 at 4:33
  • $\begingroup$ @Esha The magnitude of $\mathbf{r}_u \times \mathbf{r}_v$ at each $(u_0,v_0)$ is intuitively the ratio between the area of a small region $R$ on the surface near $\mathbf{r}(u_0,v_0)$ and the area of the region in the $(u,v)$ plane that maps into $R$. You can think of it as a generalization of the Jacobian determinant for changes of variable in 2D integrals; the Jacobian determinant is what falls out of evaluating $\| \mathbf{r}_u \times \mathbf{r}_v \|$ when $\mathbf{r}$ parametrizes a flat surface. $\endgroup$ Commented Aug 24, 2020 at 4:35
  • $\begingroup$ Does that mean the Jacobian is a generalized form of the normal vectors to be used in different scenarios? $\endgroup$ Commented Aug 24, 2020 at 4:41
  • $\begingroup$ @Esha It is really the other way around. You can think of a change of variable in 2D integrals as a special case of parametrizing a surface integral in the case when the surface you want to parametrize is flat. The Jacobian determinant that you get from the change of variable, with the absolute value, is exactly $\| \mathbf{r}_u \times \mathbf{r}_v \|$. But with this cross product method, we can also deal with curved surfaces in three dimensional space. $\endgroup$ Commented Aug 24, 2020 at 4:45
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    $\begingroup$ You can parameterize the surface $z=f(x,y)$ with the standard parameterization $$\vec{r}(u,v)=(u,v,f(u,v))$$ It turns out that $$dS=||\vec{r}_u\times\vec{r}_v||dA=\sqrt{1+(f_x(u,v))^2+(f_y(u,v))^2}dA$$ But this formula is only valid for the parameterization we just developed. In fact, if we parameterized the surface $z=f(x,y)$ by $$\vec{r}(u,v)=\Big(u^3,v^5,f\big(u^3,v^5\big)\Big)$$ then we would get a totally different surface area component $dS$. In other words, $dS$ is completely determined by how you parameterize the surface, and this may or may not equal the modulus of some gradient. $\endgroup$ Commented Aug 24, 2020 at 6:04
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This formula is for a scalar values function f.

enter image description here

And this function is for a vector valued function f.

enter image description here

The example then goes on to discuss a problem with a scalar valued function. There is no need to need to calculate a normal vector in this case.

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  • $\begingroup$ Within the method that they are using, you find a normal vector in order to find the area scale factor $\frac{dS}{dA}=\| \mathbf{r}_u \times \mathbf{r}_v \|$, even when integrating a scalar valued function on the surface. This is the usual way it's done in calculus textbooks. $\endgroup$ Commented Aug 25, 2020 at 12:30
  • $\begingroup$ $(\mathbf r_u\times \mathbf r_v)$ does generate a normal vector. And $\|(\mathbf r_u\times \mathbf r_v)\|$ generates the scaling factor. The $\frac {\mathbf n}{\|\mathbf n\|}$ factor can be distraction in some ways, as $(\mathbf r_u\times \mathbf r_v) = \frac {\mathbf n}{\|\mathbf n\|}\|(\mathbf r_u\times \mathbf r_v)\|$ $\endgroup$ Commented Aug 25, 2020 at 16:58
  • $\begingroup$ Yeah, this has been a point of confusion when I have taught this subject before: technically, you can do a flux integral by finding the unit normal $\mathbf{n}$, dotting the vector field with it, and integrating the resulting scalar field $dS$...but if you're doing the calculation by parametrizing the surface, you wind up dividing by $\| \mathbf{r}_u \times \mathbf{r}_v \|$ to normalize the normal vector and then multiplying by it again to incorporate $dS/dA$. $\endgroup$ Commented Aug 25, 2020 at 21:12
  • $\begingroup$ (Cont.) That's all technically correct, and provides a theoretical connection between the two, but it's easier to do calculations by thinking of them as distinct. Anyway, the point of my previous comment was that even when doing a surface integral of a scalar field, if you're using parametrization, you still typically construct a normal vector anyway, even though its direction is irrelevant. $\endgroup$ Commented Aug 25, 2020 at 21:13

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