Let E be a convex shape in 2 dimensions. The line integral expressed as $\int_{\partial E}f(x,y)ds$. Let $E’=\{(a*x,y/a) : (x,y) \in E \}$?
How can I transform $\int_{\partial E’}f(x,y)ds$ to a line integral on E.
I was thinking $\int_{\partial E}f(a*x,y/a)*\sqrt{a^2+1/a^2}ds$ but that trivially does not work.
This is a problem with no simple answer. Perhaps you can be more explicit about the nature of $E$, but you should not expect any universal answer here.
The "fudge factor" needs to be a (non-explicit) function of $x$ and $y$. If we parametrize $\partial E$ by $\gamma(t)=(x(t),y(t))$, then $\tilde\gamma(t)=(ax(t),y(t)/a)$ parametrizes $\partial E'$. Note that \begin{align*} ds &= \|\gamma'(t)\|\,dt = \sqrt{x'(t)^2+y'(t)^2}\,dt \\ d\tilde s &= \|\tilde\gamma'(t)\|\,dt = \sqrt{a^2x'(t)^2+y'(t)^2/a^2}\,dt = \frac{\sqrt{a^2x'(t)^2+y'(t)^2/a^2}}{\sqrt{x'(t)^2+y'(t)^2}}\,ds. \end{align*}
You can do a little better if you write the arclength integral as a typical "work" integral with $dx$ and $dy$. If $\vec n=(n_1,n_2)$ is the unit outward-pointing normal to $\partial E$, then it's standard that $$ds = -n_2\,dx + n_1\,dy \quad\text{and}\quad d\tilde s = -\tilde n_2\,dx+\tilde n_1\,dy.$$ The mapping $\phi(x,y) = (ax,y/a)$ will map a tangent vector $\vec v = (v_1,v_2)$ at a point of $\partial E$ to $\vec{\tilde v} = (av_1,v_2/a)$ at the corresponding point of $\partial E'$. Thus $\tilde n = (n_1/a,an_2)$ is normal at the corresponding point of $E'$, but it is no longer necessarily of unit length. As you can see, this is getting messier and messier: $$\tilde n = \frac {(n_1/a,an_2)}{\sqrt{(n_1/a)^2 + (an_2)^2}}.$$ So, pulling back the $d\tilde s$ integral will result in something like $$\int_{\partial E} f(ax,y/a)\frac{-an_2\,(a\,dx) + \frac{n_1}a\,(\frac{dy}a)}{\sqrt{(n_1/a)^2 + (an_2)^2}}.$$ This is no obvious multiple whatsoever of the integrand $f(ax,y/a)\,ds$.