# Elastic potential energy per unit volume,stress-strain and strain-stress curve

From what I understand, when we calculate elastic potential energy per unit volume of a material which extends linearly, we calculate the area under the graph of stress- strain OR strain- stress graph, they both will give the same value. Likewise, if I want to calculate the elastic potential energy of a linear extension, I could get the value from the area under the graph of force-extension/ extension- force graph. However, I am unable to understand why we can’t do the same for a non linear extension. For example in a non linear extension, the areas obtained from stress-strain and strain-stress curve are not the same. I only know that this has something to do with integration, but I still can’t seem to wrap my head around it. So to conclude, my question is 1) why we can’t obtain elastic potential energy per unit volume under a curve of a strain stress graph

If $$F=kx$$, in which $$k$$ is a constant, $$\int_{x=0}^X Fdx=\int_{x=0}^X kxdx=\left[\tfrac12\ kx^2\right]^X_0=\tfrac12F_XX$$ in which $$F_X=kX$$. But $$\int_{F=0}^{F_X} x dF=\int_{x=0}^X x d(kx)=\int_{x=0}^X k x dx=\tfrac12F_XX$$ This equality of the two integrals is wholly dependent on our initial assumption: $$F=kx.$$ The same argument applies for the stress vs. strain graph.
Less precisely, if the graph of $$F$$ against $$x$$ started as $$F=kx$$ but curved upwards for higher values of $$x$$, this would make the area between the graph and the x-axis (up to a given value, $$X$$) higher, but would make the area between the graph and the $$F$$ axis (up to $$F_X$$) lower!
• The area between the graph line and the $x$ axis. This follows from the definition of $work.$ – Philip Wood Jun 2 at 7:49