Double Integral over triangular region

Question

$\displaystyle\int\int_{R}x\,\mathrm{d}A$ over the triangular region R enclosed by the $x + 2y =2$, $x = 0$, and $y = 0$.

I need help to solve it. Thanks in advance.

Answer 1

$\int x\ dA$ will give you the $x$ coordinate of the centroid of A

A has vertexes $(2,0),(0,1),(0,0)$ The centroid is $(\frac 23, \frac 13)$

$\frac 23$

You want to learn how to set up the limits of integration?

I suggest you always start with a sketch of your region. Draw a vertical (or horizontal) line through the your sketch of the region.

How would you describe the endpoints of that line as equations in terms of x.

One end is $(x,0)$ the other end is $(x,1 - \frac 12 x)$

This gives us our limits for y.

$\int (\int_0^{1-\frac 12 x} x\ dy) \ dx$

And what are our limits for $x?$ The biggest $x$ can be is $2$ the smallest is $0.$

$\int_0^2 (\int_0^{1-\frac 12 x} x\ dy) \ dx$

If you want to change the order of integration use the horizontal line instead. Find the equations, in terms of y, that describe the endpoints, and find the total range of $y.$

$\int_0^1 (\int_0^{2-2y} x \ dx)\ dy$