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

[Amit kumar sahu]

 

[LCKurtz]

Try polar coordinates and show us what you get.

 

[HallsofIvy]

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 $\displaystyle \pi/2$:
$\displaystyle \int_{r= 0}^a\int_{\theta= 0}^{\pi/2} (r cos(\theta))(r sin(\theta))(r drd\theta)$
$\displaystyle = \int_{r=0}^a\int_{\theta= 0}^{\pi/2} r^3 cos(\theta) sin(\theta) drd\theta$
$\displaystyle = \left(\int_0^a r^3 dr\right)\left(\int_0^{\pi/2}cos(\theta)sin(\theta)d\theta\right)$.

 

[Amit kumar sahu]

Thanks ❤