Q$1)$ If the Airys stress function $$ is defined in cartensian coordinates such that; ${}_x=\frac{de{l}^2}{del{y}^2}$;$,{}_y=\frac{de{l}^2}{del{x}^2}$;$,{}_{xy}=-\frac{de{l}^2}{\text{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}},$ using transformation equations $r^2=x^2+y^2$ and $tan=\frac{y}{x}$ show that; $\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{x}}=\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{r}}cos-\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}}\frac{sin}{r}$ and $\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{y}}=\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{r}}sin+\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}}\frac{cos}{r}$ Hence find stress components in polar coordinates system.