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If

$\int_a^b f(x)-g(x)dx$

is the formula for the area between curves, what happens if we have $ f(x,y) $? Is there any way to calculate the area between curves of multi-variable functions? Thank you.

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Yes. Double integrals tend to occupy about 2-3 weeks of 2nd-year calculus courses. It takes students longer than that to get used to them.

Here, instead of an interval $[a,b]$, you are integrating over a region $R$, so:

$$\iint_R[f(x,y)-g(x,y)]dA$$

If you can describe the region as being bounded between $y=c(x)$ and $y=d(x)$, with $x\in [a,b]$, you could convert this to a pair of single integrals:

$$\int_a^b\left[\int_{c(x)}^{d(x)}[f(x,y)-g(x,y)]dy\right]dx.$$

For example, if $f=x^2+y^2$ and $g=0$, and $R$ is the triangle between (0,0), (2,0) and (2,2), then $a=0$, $b=2$, $c(x)=0$ and $d(x)=x$, and the double integral $$\iint_R(x^2+y^2)dA$$ is $$\int_0^2\left[\int_0^x(x^2+y^2)dy\right]dx$$ which is $$\int_0^2\left[x^2y+\frac{y^3}{3}\right]_0^xdx,$$ that is, $$\int_0^2\left[x^3+\frac{x^3}{3}\right]dx,$$ which is $$\left[\frac{x^4}4+\frac{x^4}{12}\right]_0^2,$$ which is $\frac{16}{4}+\frac{16}{12}=\frac{16}{3}$.

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Let's rewrite the given formula to: $$∫_{a}^b f(x) -g(x)\mathrm{d}x = ∫_{a}^b f(x) \mathrm{d}x - ∫_{a}^bg(x)\mathrm{d}x.$$

As you see, that formula simply subtracts the two areas below the curve. There is nothing hidden in the formula.

The graph of $f:ℝ×ℝ→ℝ, (x,y)↦z$ is not a curve, but a surface. So the 'area' below that surface is not an area, but a volume.

And again you can do: $$∫_{Ω} f(x,y) \mathrm{d}(x,y) - ∫_{Ω}g(x,y)\mathrm{d}(x,y) = ∫_{Ω} f(x,y) - g(x,y)\mathrm{d}(x,y),$$ with some given region $Ω∈ℝ²$.

I found some nice illustrations and examples about 2D-integrals here.

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  • $\begingroup$ small remark: It is more difficult to actually calculate that expression in 2D. For the 1D case, you have to determine the points, where both curves intersect. In 2D that is not easy. $\endgroup$ – P. Siehr Jul 25 '17 at 7:57

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