# How does $\cos\alpha dA=dydz$ come?

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.

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.

• How $dA$ is the are spanned by $\vec{r_u}du$ and $\vec{r_v}dv?$ – Math geek Mar 10 at 10:24
• In the fifth line, Isn't it $dy dz$ instead of $dx dy$? – Math geek Mar 10 at 10:26
• What is $P_{yz}$? – Math geek Mar 10 at 10:27
• Can you explain bit? – Math geek Mar 10 at 10:28
• 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. – setnoset Mar 10 at 10:49