A multivariable integral problem im solving for first time , please help me with a solution

The "positive quadrant" is the first quadrant where both x and y are positive. That is, of course, a quarter of the circle \(\displaystyle x^2+ y^2= a^2\).
We can solve that for y as \(\displaystyle y= \sqrt{a^2- x^2}\) or for x as \(\displaystyle x= \sqrt{a^2- y^2}\).

One way to do this is to take x going from 0 to a and, for each x, take y going from 0 to \(\displaystyle \sqrt{a^2- x^2}\):
\(\displaystyle \int_{x= 0}^a \int_{y= 0}^{\sqrt{a^2- x^2}} xy dydx\).

Another way to do this is to take y going from 0 to a and, for each y, take x going from 0 to \(\displaystyle \sqrt{a^2- y^2}\):
\(\displaystyle \int_{y= 0}^a \int_{x= 0}^{\sqrt{a^2- y^2}} xy dxdy\).

Given the circular symmetry, the simplest is, as LCKurtz suggests, using polar coordiates. \(\displaystyle x= rcos(\theta)\), \(\displaystyle y= r sin(\theta)\), \(\displaystyle dxdy= r drd\theta\), r goes from 0 to a and \(\displaystyle \theta\) goes from 0 to \(\displaystyle \pi/2\):
\(\displaystyle \int_{r= 0}^a\int_{\theta= 0}^{\pi/2} (r cos(\theta))(r sin(\theta))(r drd\theta)\)
\(\displaystyle = \int_{r=0}^a\int_{\theta= 0}^{\pi/2} r^3 cos(\theta) sin(\theta) drd\theta\)
\(\displaystyle = \left(\int_0^a r^3 dr\right)\left(\int_0^{\pi/2}cos(\theta)sin(\theta)d\theta\right)\).
 
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