I suppose the integral is
[tex]\displaystyle\int_0^5\int_y^{\sqrt{25-y^2}} xy\,\mathrm dx\,\mathrm dy[/tex]
The integration region corresponds to a sector of a cirlce with radius 5 subtended by a central angle of π/4 rad. We can capture this region in polar coordinates by the set
[tex]R=\left\{(r,\theta)\mid0\le r\le5,0\le\theta\le\dfrac\pi4\right\}[/tex]
Then [tex]x=r\cos\theta[/tex], [tex]y=r\sin\theta[/tex], and [tex]\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta[/tex]. So the integral becomes
[tex]\displaystyle\iint_Rxy\,\mathrm dx\,\mathrm dy=\int_0^{\pi/4}\int_0^5r^3\sin\theta\cos\theta\,\mathrm dr\,\mathrm d\theta=\frac{625}{16}[/tex]