Setting up and evaluating the double integral for the function $f(x,y) = x^2y$ over the region $T$

Question

I am having trouble setting this up, can someone please help me?

Let $T$ be the triangle in the $x,y$ plane having corners at $(0,0), (1,3),$ and $(3,3)$. Can someone please help me set up the double integral for the function $f(x,y) = x^2y$ over the region $T$, and evaluate it, please. Thank you!

Answer 1

You need to look at the region and determine the bounds. If we graph out the points and draw lines, it becomes clear that using functions of $y$ will make this integral simpler since we won't need to split it into two regions. First figure out what the equations of our three lines are. The top line is easy:

$$y=3$$ The Left most line is: $$y=3x\Rightarrow x=\frac{1}{3}y$$ The Right most line is: $$y=x\Rightarrow x=y$$

We can now define our region $T$ as follows:

$$T=\{(x,y)\in\mathbb{R}^2|\frac{1}{3}y\leq x \leq y,0\leq y\leq3\}$$

From here we can set up our integral:

$$\iint_Tx^2ydA=\int_0^3\int_{\frac{1}{3}y}^yx^2ydxdy$$ $$=\frac{1}{3}\int_0^3[yx^3]|_{\frac{1}{3}y}^y$$ Simplifying our integrand we obtain: $$[yx^3]|_{\frac{1}{3}y}^y=y[y^3-\frac{y^3}{27}]$$ $$=\frac{26}{27}y^4$$ Thus our integral becomes: $$\frac{26}{81}\int_0^3y^5dy=\frac{26}{81*5}[y^5]|_0^3$$ $$=\frac{26}{81*5}*243=\frac{3*26}{5}=\frac{78}{5}$$

Answer 2

Hints: $$T=T_1 \cup T_2, \space \space T_1: 0 \le x \le 1, \space x \le y \le 3x; \space\space T_2: 1 \le x \le 3, x \le y \le 3.$$

Figure:

Answer 3

Your region looks like this:

Call $R_1$ the orange region and $R_2$ the blue region. Observe that for any point $(x,y)$ in $R_1$, the $x$-coordinate is always between $0$ and $1$, and the $y$-coordinate is always below the line $y=3x$ and above the line $y=x$. So $$R_1 = \{(x,y) : 0 \leq x \leq 1, \, x \leq y \leq 3x\}.$$ Similarly, one has $R_2 = \{(x,y) : 1 \leq x \leq 3, \, x \leq y \leq 3\}$. Hence $$\int_T f = \int_{R_1}f + \int_{R_2}f = \int_0^1 \int_x^{3x} f(x,y) dydx + \int_1^3 \int_x^{3} f(x,y) dydx.$$