How do I calculate this surface integral over a surface of a triangle with given vertices?

Question

$\displaystyle\iint_{A}(2x-3y)dxdy$ where $A$ is the triangle with given vertices $(0,0),(2,1),(2,0)$

I don't know how to set up the limits in such cases. What will be the methods of setting limits of the two integrals of the variables $x$ and $y$ ? And how to set them in such cases. I need help with the process of setting limits, then, I am good with the rest.

Answer 1

You draw the triangle and find the equation of the slant side, in this case $y=\frac{x}{2}$

Thus the limits are $0\le x \le 2$ and $0\le y\le \frac{x}{2}$

So you can write the integral as $$\int_0^2 \left(\int_0^{\frac{x}{2}} (2 x-3 y) \, dy\right) \, dx$$

Hope this can be useful

$$ . $$

Answer 2

Sketch the region $A$:

Then, solve accordingly: $$\displaystyle\iint_{A}(2x-3y)dxdy = \int_0^2 \int_0^\frac{x}{2} (2x-3y) dy dx = \\ = \int_0^2 \left( 2xy - \frac{3}{2}y^2\right)_{y=0}^{y=\frac{x}{2}}dx = \\ = \int_0^2 \left( 2\frac{x^2}{2} - \frac{3}{2}\frac{x^2}{4}\right)dx = \\ = \int_0^2 \frac{5}{8}x^2dx = \frac{5}{8}\left(\frac{x^3}{3}\right)_{x=0}^{x=2} = \frac{5}{3}.$$