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

To evaluate the integral $\iint_R xy \,dx\,dy$ over the region R, we first identify the vertices of the region R. The given lines are:

• $y = x + 2$
• $y = -x + 2$
• $y = x$
• $y = -x$

The region R is a diamond centered at the origin (0,0) with vertices at:

• Intersection of $y = x + 2$ and $y = -x + 2$: Solve $x + 2 = -x + 2 \Rightarrow x = 0$, $y = 2$. Vertex: (0,2)
• Intersection of $y = x + 2$ and $y = x$: Solve $x + 2 = x \Rightarrow$ no solution here directly.
• Intersection of $y = -x + 2$ and $y = -x$: Solve $-x + 2 = -x \Rightarrow$ no solution here directly.
• Intersection of $y = x$ and $y = -x$: Solve $x = -x \Rightarrow x = 0$, $y = 0$. Vertex: (0,0)

The full vertices are (0,2), (2,0), (0,-2), and (-2,0).

Since the region is symmetric with respect to both axes, the integral of an odd function like $xy$ over it should be zero.

Consider the symmetry about the x- and y-axes: the contributions from symmetrical parts about the axes cancel each other out.

Thus, the value of the integral is 0, which is within the given expected range of 0 to 0.

Therefore, the value of the integral, rounded to the nearest integer, is 0.