4

a) Is it true that $$\iint_{R} f(x,y)dydx=\iint_{R} f(x,y)dxdy$$

I thought this would be true because of Fubini's Theorem however Fubini's Theorem requires $f(x,y)$ to be continuous on $R$. There is no such condition in this problem. Therefore it is false?

b) If $R$ is the rectangle $0\leq x\leq a, 0\leq y\leq b$ and $S$ is the rectangle $-a\leq x\leq 0, -b \leq y\leq 0$ then

$$\iint_{R} f(x,y)dA=-\iint_{S} f(x,y) dA$$

I think this is false and the counter example I would use is

$f(x,y)=1$

Then $\int_{0}^{b}\int_{0}^{a} dydx = \int_{-b}^{0}\int_{-a}^{0}dydx$

Is this correct?

Another User
  • 5,048
user130306
  • 1,890

1 Answers1

1

a) Is it true that $$\iint_{R} f(x,y)dydx=\iint_{R} f(x,y)dxdy$$

I thought this would be true because of Fubini's Theorem however Fubini's Theorem requires $f(x,y)$ to be continuous on $R$.

The continuity version of Fubini's Theorem in $\mathbb R^2$ is a special case.

The above equality does hold as long as $\int f(x,y)\,\mathrm dx$ and $\int f(x,y)\,\mathrm dy$ both exist on rectangle $R$'s respective edges.

b) If $R$ is the rectangle $0\leq x\leq a, 0\leq y\leq b$ and $S$ is the rectangle $-a\leq x\leq 0, -b \leq y\leq 0$ then

$$\iint_{R} f(x,y)dA=-\iint_{S} f(x,y) dA$$

I think this is false and the counter example I would use is

$\int_{0}^{b}\int_{0}^{a} dydx = \int_{-b}^{0}\int_{-a}^{0}dydx$

Is this correct?

Yes (ignoring your typo where you interchanged $x$ and $y).$

ryang
  • 38,879
  • 14
  • 81
  • 179