Does this double integral give area or volume?

Question

I've encountered a question like this: $$\int_{-1}^0 \int_0^{-1}x y\ dy \ dx.$$ I think it represents the volume of some solid object in the 3rd quadrant and the value of f(x,y) is always positive as both the x-coordinate and the y-coordinate are negative and their product will always be positive. So we can assume that the volume of the solid is along the positive z-axis, so positive volume. Then why is the answer -0.25, which is negative?

Answer 1

The integral is indeed computing a volume between the function values of the integrand $xy$ in a domain bounded by the integral boundaries.

The value is negative, due to the "direction" you are integrating over the $y$-axis. Flip it and you get the same answer with a positive sign:

$$\int_{-1}^0 \int_{-1}^0x y\ dy \ dx.$$

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Edit:

The sign of the value of the integral is related to the integration direction, due to the rules of integration you are applying. In this specific case, we can compute a simpler integral of the function $f(x)=x$ step by step, to observe this. $$\begin{align} \int_{0}^1 x dx &= \left. \frac{x^2}{2} \right|_0^1=\frac{1}{2}-\frac{0}{2} = \frac{1}{2}\\ \int_{1}^0 x dx &= \left. \frac{x^2}{2} \right|_1^0=\frac{0}{2}-\frac{1}{2} = -\frac{1}{2} \end{align}$$

You see, the function we integrate doesn change, only the direction, which flips the sign of the answer.

If you, however, are interested in the area $A[f]$ under the curve of the function, then you should apply the absolute value:

$$A[f] = \left\vert \int_{a}^b x dx \right\vert = ...$$

This result will always be positive, as we would expect from an area (or volume).

Answer 2

Think of the double integral like this:

Visualise a 3-D plot of the function $f(x, y) = xy$ on $\mathbb R^3$. It should look somewhat like this:

(you can use this link if you want to try it out yourself)
Consider an infinitesimally small piece, called $A_0$, of the X-Y plane of area $dA = dx \cdot dy$. Then, the SIGNED volume under the curve over that small piece of area will be $dV = \displaystyle \int\int_{A_0}xy\ dx\ dy$. Summing the little volumes up, taking areas which span from the given limits of x and y, will give us the required sums of SIGNED volumes, $\displaystyle \int_{y_i}^{y_f}\int_{x_i}^{x_f}xy\ dx\ dy$.

NOTE: The signed volume means that if the curve is under the XY plane, i.e. $f(x, y) < 0$, then $dV < 0$. The direction in which you integrate matters, as you can see because the function seems positive on $[-1, 0]\times[-1,0]$. But the conventional direction of increment is in the right and upward directions, which have been flipped in the question, thus making the integral negative.