How to prove that the area of a shape is independent of the choice of axes? Suppose we have some shape in a plane, and we want to find its area using calculus.
I set my $x$ and $y$ axes arbitrarily. They are perpendicular to each other.
I can calculate the area of the region using a standard double integral, slicing up the area into rectangles with sides parallel to the $x$ and $y$ axes.
$$A = \int_{x_1}^{x_2}\left(\int_{y_1(x)}^{y_2(x)}dy\right)dx$$
Now if I rotate my axes (still remaining perpendicular to each other) but leave the shape unchanged, and calculate the integral the same way but with the new $x$ and $y$ axes and new $x$ and $y$ bounds... I'm looking for a proof that the area will turn out the same.
 A: Transformations that preserve distances in a metric space are called isometries and the only isometries of Euclidean spaces are rotations, translations, reflections or some composition of those operations. Rotations and reflections are non-trivial but thankfully they're easily characterized as they're linear transformations. Conceptually this proof relies on the geometric interpretation of the determinant as a signed volume of a parallellpiped and the fact that distance preserving transformations will also preserve areas, just like in your integral. This means they'll have determinant of $\pm 1$. All that's left to prove is that every such matrix preserves distances. Lets get started.
First, we'll use column vectors and for two vectors $x$ and $y$ we'll use matrix multiplication to represent the inner product $\langle x,y \rangle = x^Ty$ with $x^T$ the transpose of $x$. This allows us to write the Euclidean metric as $||x|| = \sqrt{x^Tx}$. Now we're interested in matrices that act on $x$ by multiplication but preserve distances, that is to say we're interested in an an $A$ such that $||Ax||=||x||$ and by squaring both sides we can simplify this to $(Ax)^TAx = x^Tx$. Since $(Ax)^T = x^TA^T$ this implies that $x^TA^TAx=x^Tx$ and so we must have that $A^TA=I$, the identity matrix. This means $A$ must be an orthogonal matrix and in particular, has determinant $\pm 1$ because $\det A = \det A^T$ and $\det(A^TA) = \det A^T \det A$ we have that $\det A^TA = \det I = 1$ and so $\det A = \pm 1$. Since every matrix of this form has this property they are all isometries and we're done. Note that I never had to mention the number of dimensions either so this is the case for $\mathbb{R}^n$. If you want to work over the complex numbers you replace the transpose with the conjugate transpose but the ideas are the same.
A: I see that this can be proven using Green's Theorem.
https://tutorial.math.lamar.edu/classes/calciii/GreensTheorem.aspx
So we can write the area of a region as a line integral around the boundary:
$$A = \oint_C xdy = -\oint_C ydx$$
So if we rotate our axes clockwise by $\theta$ radians:
$$x_{new} = cos(\theta)x - sin(\theta)y$$
$$dy_{new} = sin(\theta)dx + cos(\theta)dy$$
$$A_{new} = \oint_C x_{new}dy_{new} $$
$$A_{new} = \oint_C (cos(\theta)x - sin(\theta)y)(sin(\theta)dx + cos(\theta)dy) $$
$$A_{new} = \left(cos(\theta)sin(\theta)\oint_C xdx\right) + \left(cos^2(\theta)\oint_C xdy\right) - \left(sin^2(\theta)\oint_C ydx\right)- \left(sin(\theta)cos(\theta)\oint_C ydy\right)$$
First and last integrals go to 0.
$$A_{new} = \left(cos^2(\theta)\oint_C xdy\right) - \left(sin^2(\theta)\oint_C ydx\right)$$
$$A_{new} = \left(cos^2(\theta)+sin^2(\theta)\right)A$$
$$A_{new} = A$$
Green's Theorem also shows that an area can be divided arbitrarily into subregions. The sum of the small line integrals adds up to a line integral around the whole region (edges being shared by regions cancel each other out). This demonstrates that the area integral is additive.
I would like to see other answers as this feels like overkill to use Green's Theorem prove a basic idea about integrals and areas.
A: The result you want is that if $E$ is a "shape in the plane", then $A(E)= A(R(E)+ v),$ where $R$ is a rotation, and $v\in \mathbb R^2.$ In other words, area is an invariant under rotations and translations. The problem you're having may be due to problems with defining "area" rigorously. The ideas in Lebesgue measure theory make short work of this, once you have it, but of course you are probably not in a measure theory course.
A: Just use integration by substitution ($u$-substitution).

*

*Two things to note about rotations. First, every rotation transformation $\mathbf{R}_\theta(x,y)$ has a matrix determinant of 1 for all $x$ and $y$. You can prove this to yourself using:
$$\mathbf{R}_\theta(x,y) = \begin{bmatrix}\cos{\theta}&\sin{\theta}\\-\sin{\theta}&\cos{\theta}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\\=\left\langle x\cos{\theta}+ y\sin{\theta},\;-x\sin{\theta}+y\cos{\theta}\right\rangle$$
Second, the total derivative of every rotation is itself. You can prove this using the above equation; it is true in general because a rotation is a linear coordinate transform. You can represent it as a matrix multiplication.


*The substitution formula says that if $\varphi(\mathbf{u})$ is an invertible and appropriately smooth transformation in some region $U$, then:
$$\int_{\varphi(U)} f(\mathbf{v})d\mathbf{v} = \int_{U} f(\varphi(\mathbf{u}))\cdot\left|\det(D\varphi)(\mathbf{u})\right| d\mathbf{u}$$


*For your particular case, you have a shape $S\subset \mathbb{R}^2$ whose integral you want to calculate. You have some rotation $\mathbf{R}_\theta$ which transforms $S$ into a rotated version of itself $\widehat S$. Because we are just integrating to get area, let $f$ denote the (boring) constant function $f(x)=1$. Applying the variable substitution formula,
$$\begin{align*}\iint_{\widehat S} dx^\prime dy^\prime &= \iint_{\mathbf{R}_\theta(S)} f(x^\prime,\,y^\prime) dx^\prime dy^\prime\\&=\iint_S f\left(\mathbf{R}(x,y)\right) \cdot\left|\det(D\mathbf{R})(x,y)\right|dx\,dy\\  
&=\iint_S 1 \cdot\left|\det(D\mathbf{R})(x,y)\right|dx\,dy & \{f \text{ is constant}\}\\ 
&=\iint_S 1 \cdot\left|\det(\mathbf{R})(x,y)\right|dx\,dy & \{\text{rotation is linear}\}\\ 
&=\iint_S 1 \cdot\left|1\right|dx\,dy & \{\det=1\}\\ 
&=\iint_S \; dx\, dy
\end{align*} $$

*

*We have proven that the area will be the same before and after. The two key features we used were that (1) the derivative of a linear transformation (such as a rotation) is equal to the function itself: $D_{\mathbf{x}}(\mathbf{A}\cdot \mathbf{x}) = \mathbf{A}$, and (2) the Jacobian determiant of a rotation is equal to 1 for every $x$ and $y$. (This is really the key part; the Jacobian determinant tells you how the infinitesimal area of $dx\,dy$ is scaled up by the transformation. If the Jacobian were 2 everywhere, it would double the area. If the Jacobian varied with $x$ and $y$, the transform might shrink some areas and grow others. Because it's 1 everywhere, the infinitesimal area remains unaltered at all points.)


*Note that we can use this same argument to show that reflections preserve area as well; they're linear transformations with a determinant of -1. Thanks to the absolute value signs in the variable substitution formula, the same reasoning applies.  Similarly for a glide transform, which is a reflection followed by a rotation.
