The area enclosed by a space curve in $\mathbb{R}^3$ after projected onto the plane

Question

Let $\mathbf{r}(t)$ be a space curve in a Euclidean space $\mathbb{R}^3$. In class, I learned that $\hat{i} \cdot \int \mathbf{r} \times d\mathbf{r}$ (where $\cdot$/$\times$ isa dot/cross product) is proportional to the area enclosed by the curve $\mathbf{r}$ after it is projected onto the plane orthogonal to $\hat{i} \in \{x, y, z\}$. I didn't understand why this is true. Can anyone help me to understand?

Answer 1

One way to understand this is to denote $C = \{\mathbf{r}(t) : a\leq t\leq b\} $ and recognize that

$$ \hat{i}\cdot \int_C \mathbf{r}\times d\mathbf{r} = \int_a^b \hat{i}\cdot\mathbf{r}(t)\times\mathbf{r}'(t)\,dt = \int_a^b \mathrm{det}\left(\hat{i},\mathbf{r}(t),\mathbf{r}'(t)\right)dt, $$

so that for each $t$ the integrand measures the oriented volume spanned by the vectors $\hat{i}, \mathbf{r}(t)$ and $\mathbf{r}'(t)$.

By the properties of the determinant, this quantity relies only on the component of $\mathbf{r}(t)\times\mathbf{r}'(t)$ in the direction of $\hat{i}$ (check this as an exercise), let’s call this number $c$. Since $|\mathbf{r}\times \mathbf{r}’|$ is the area of the parallelogram spanned by $\mathbf{r}$ and $\mathbf{r}’$ it follows that $c$ is the area of the projection of this parallelogram onto the plane normal to $\hat{i}$. Therefore, the volume density at each point expressed by the original integral is proportional to the area density in this plane, which is orthogonal to $\hat{i}$ as just discussed.

This is why the result of the original integral will be proportional to the area enclosed by the projection of $\mathbf{r}$ onto the plane orthogonal to $\hat{i}$: the integral simply sums up these local contributions.

EDIT: You asked for an example, so consider the following simple one. Let $\mathbf{r}(t) = (\sin t, \cos t, t)$ from $0\leq t\leq 2\pi$ be a helix aligned with the $\mathbf{e}_3$-direction. Then, choosing $\hat{i}=\mathbf{e}_3$, the theorem says that the integral above should be proportional to the area enclosed by the projection of this helix onto the plane normal to $\mathbf{e}_3$, which is the area of a circle of radius 1, i.e., $\pi$. Let’s check that this is the case.

We have $\mathbf{r}’(t) = (\cos t, -\sin t, 1)$, and so

$$ \hat{i}\cdot \int_C \mathbf{r}\times d\mathbf{r} = \int_0^{2\pi}(\cos^2t+\sin^2t)dt = 2\pi \sim \pi,$$

confirming that the value of the integral is proportional to the area of the projection. Note that the constant of proportionality depends on the number of “winds” in the original helix, which cannot be determined from its shadow on the relevant plane.