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In the red rectangle, author defined what is surface integral in terms of parametric form. I am confused with the expression in the yellow rectangle. Can you please explain? How does $$\cos\alpha\; dA=dy\,dz$$ come? For the rest of the expression, I can do it by using that. Please help me.

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Take $\alpha$ for instance. That is defined by the author to be the angle between the x axis and $\mathbf{n}$, so that $\cos\alpha = \mathbf{n}\cdot\mathbf{i}$, that is, the x component of the unit normal. Now $dA$ is the area of the parallelogram spanned by two vectors $\mathbf{r_u}du$ and $\mathbf{r_v}dv$. What we want is to project these vectors into the yz plane which will give us the $dydz$ area element:

\begin{align} dy\ dz &= \mathbf{i} \cdot \lbrack P_{yz}(\mathbf{r_u}du) \times P_{yz}(\mathbf{r_v}dv) \rbrack \\ &= \lbrack (\mathbf{j} \cdot \mathbf{r_u})\ (\mathbf{k} \cdot \mathbf{r_v}) - (\mathbf{k} \cdot \mathbf{r_u})\ (\mathbf{j} \cdot \mathbf{r_v}) \rbrack \ du\ dv \\ &= \mathbf{i} \cdot (\mathbf{r_u}du \times \mathbf{r_v}dv) = \mathbf{i} \cdot \mathbf{N}\ du\ dv = \cos\alpha\ dA. \end{align}

Here I used the projection operator $P_{yz}(\mathbf{v}) = \mathbf{j} \cdot \mathbf{v} + \mathbf{k} \cdot \mathbf{v} $ to project vectors onto the yz plane.

The first equation comes from the Jacobi transformation from $(u,v)$ to $(y,z)$, which is essentially a change of variable performed during multivariable integration. I will provide a heuristic argument for it, however I will only consider the notion of unsigned integrals.

Consider the map $\phi: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, assumed to be invertible in the region of integration. If you take a sufficiently small square $B =(u, u+\Delta u) \times (v, v + \Delta v)$ of area $Area(B) = \Delta u\ \Delta v$, you will map it to $\phi(B)$ which will be roughly a parallelogram spanned by the vectors $\phi_u \Delta u$ and $\phi_v \Delta v$, where the subscripts denote the partial derivatives of $\phi(u, v)$. To show this, expand $\phi(u + \Delta u, v) - \phi(u, v)$ and $\phi(u, v + \Delta v) - \phi(u, v)$ as a taylor series centered at $(u,v)$ and neglect higher order terms. Then the (signed) area of the mapped square will be: $$ Area(\phi(B)) \approx \det\left( \frac{\partial\phi}{\partial u}\ \frac{\partial\phi}{\partial v} \right) Area(B).$$

The factor $J = \det\left( \phi_u\ \phi_v \right)$ is called the Jacobian of the transformation $\phi$. I've allowed the area to be signed, which means that if the jacobian is negative, $Area(\phi(B))$ and $Area(B)$ have opposite sign. The actual value of the area is the absolute value of the unsigned area. If we define $\Delta y$ and $\Delta z$ to be the width and height of the parallelogram respectively, we find that $ \Delta y \Delta z \approx \left|J\right| \Delta u \Delta v$.

Now let $f$ be a function of two variables and let $\phi(R)$ be the region to be integrated over. Then we can approximate the signed integral (the approximation becoming exact within a limit procedure) by splitting the region into multiple boxes as follows:

\begin{align} \int_{\phi(R)}{f(y, z)\ dy\ dz} &\approx \sum_{B}{f(y_B, z_B)}Area(\phi(B)) \\ &\approx \sum_{B}{f(\phi(u_B, v_B))\ J Area(B)} \\ &\approx \int_{R}{f\circ\phi(u, v) \ J \ du\ dv} \end{align}

Returning to the initial formula, if we project the parallelogram spanned by $\mathbf{r_u}du$ and $\mathbf{r_v}dv$ onto the xy plane, we actually get a parallelogram spanned by the projected vectors $P_{yz}(\mathbf{r_u}du)$ and $P_{yz}(\mathbf{r_v}dv)$. Suppose that we can express our surface as a function of the $x$ coordinate, i.e. $x = \sigma(y, z)$, then we may define $f(y, z) = F_1(\sigma(y, z), y, z)$ and the map $\phi(u, v) = P_{yz}(\mathbf{r})$. Then it follows that the Jacobian of the transformation is $J = \mathbf{i} \cdot \lbrack P_{yz}(\mathbf{r_u}du) \times P_{yz}(\mathbf{r_v}dv) \rbrack$. We thus obtain our original equation. An actual rigorous proof is probably best studied in the context of differential forms, where the integrands of surface integrals are naturally expressed as 2-forms. See these notes on multivariable calculus by Jerry Shurman for more information.

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  • $\begingroup$ How $dA$ is the are spanned by $\vec{r_u}du$ and $\vec{r_v}dv?$ $\endgroup$ – Math geek Mar 10 at 10:24
  • $\begingroup$ In the fifth line, Isn't it $dy dz$ instead of $dx dy$? $\endgroup$ – Math geek Mar 10 at 10:26
  • $\begingroup$ What is $P_{yz}$? $\endgroup$ – Math geek Mar 10 at 10:27
  • $\begingroup$ Can you explain bit? $\endgroup$ – Math geek Mar 10 at 10:28
  • $\begingroup$ dA is not spanned by $\mathbf{r_v}dv$ and $\mathbf{r_v}dv$, it is the magnitude of the vector $mathbf{N} du dv$ that is perpendicular to both, hence the cross product. $\endgroup$ – setnoset Mar 10 at 10:49

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