Find total charge of triangular region

Question

Charge is distributed over a triangular region in the -plane bounded by the -axis and the lines =5− and =1+. The charge density at a point (,) is given by (,)=+, measured in coulombs per square meter (C/m2). Find the total charge.

I've graphed a triangle which then made me form the integral $$\int_0^3 \int_{y-1}^{5-y} (x+y) \,dx \, dy$$, making me get $27$. I believe there is something wrong with my calculation. Is it that I am using the wrong variable or something else?

Answer 1

You are considering the region bound by the two given lines and x-axis. But the question says y-axis.

See the diagram for the correct region.

So the integral should be,

$ \displaystyle \int_0^2 \int_{x+1}^{5-x} (x + y) ~ dy ~ dx$

Answer 2

The intersection of $5-x$ and $1+x$ is at $(2,3)$; there should be $x$ and not $y$ in the inner integral as well, leading to $$\int_0^2\int_{1+x}^{5-x}(x+y)\,dy\,dx=\frac{44}3$$