Question

The value of the integral $\iint_R xy \,dx\,dy$ over the region R, given in the figure, is ___________ (rounded off to the nearest integer). 

 

Hint:

Before performing a lengthy integration, always check for symmetry. If the region of integration is symmetric with respect to an axis (e.g., the y-axis) and the integrand is an odd function with respect to the corresponding variable (e.g., $f(-x,y) = -f(x,y)$), the integral is zero.

Answer 1

Correct Answer: 0

The region of integration R is a square defined by the lines: $y = x \implies y-x=0$ $y = -x \implies y+x=0$ $y = -x+2 \implies y+x=2$ $y = x+2 \implies y-x=2$
This region is symmetric with respect to the y-axis.
Let's check the symmetry of the integrand $f(x,y) = xy$. We test for symmetry about the y-axis by replacing $x$ with $-x$. $f(-x, y) = (-x)y = -xy = -f(x,y)$. The integrand is an odd function with respect to $x$.
When integrating an odd function over a symmetric interval, the result is zero. Since the region R is symmetric about the y-axis (if $(x,y)$ is in R, then $(-x,y)$ is also in R) and the function $f(x,y)=xy$ is odd with respect to x, the value of the double integral is zero.
$\iint_R xy \,dx\,dy = 0$.
Alternatively, we can use a change of variables. Let $u = y-x$ and $v = y+x$. The limits of integration become $0 \le u \le 2$ and $0 \le v \le 2$. From the transformation, $x = (v-u)/2$ and $y=(v+u)/2$. The Jacobian is $|J| = 1/2$. The integral becomes: $I = \int_0^2 \int_0^2 \left(\frac{v-u}{2}\right)\left(\frac{v+u}{2}\right) \left|\frac{1}{2}\right| du\,dv = \frac{1}{8} \int_0^2 \int_0^2 (v^2 - u^2) du\,dv$. $I = \frac{1}{8} \int_0^2 \left[ v^2u - \frac{u^3}{3} \right]_0^2 dv = \frac{1}{8} \int_0^2 \left( 2v^2 - \frac{8}{3} \right) dv$. $I = \frac{1}{8} \left[ \frac{2v^3}{3} - \frac{8v}{3} \right]_0^2 = \frac{1}{8} \left[ \left(\frac{16}{3} - \frac{16}{3}\right) - (0) \right] = 0$.
The value is 0, which when rounded to the nearest integer is 0.