Multiple Integrals Notes

Last reviewed 16 Sept 2026 · 5 min read

Double integrals

\iint_{R} f (x, y) d A

represents the volume under the surface $z = f (x, y)$ over region $R$ (if $f \geq 0$ ); with $f = 1$ it gives the area of $R$ .

\iint_{R} f d A = \int_{a}^{b} [\int_{c}^{d} f (x, y) d y] d x = \int_{c}^{d} [\int_{a}^{b} f (x, y) d x] d y

(Fubini's theorem — order can be interchanged for continuous $f$ .)

If $f (x, y) = g (x) h (y)$ over a rectangle: $\iint f d A = (\int g d x) (\int h d y)$ .
Type I region $a \leq x \leq b$ , $y_{1} (x) \leq y \leq y_{2} (x)$ : integrate with respect to $y$ first (inner limits are functions of $x$ ).
Type II region $c \leq y \leq d$ , $x_{1} (y) \leq x \leq x_{2} (y)$ : integrate with respect to $x$ first.

Rule: the inner integral's limits may depend on the outer variable; the outer limits must be constants.

Changing the order often makes an integral easier (or possible) to evaluate.

\iint_{R} f (x, y) d x d y = \iint_{R^{'}} f (x (u, v), y (u, v)) ∣ J ∣ d u d v, J = \frac{\partial ( x , y )}{\partial ( u , v )}

$x = r cos θ$ , $y = r sin θ$ , $∣ J ∣ = r$ :

d x d y = r d r d θ

Useful for circular regions and integrands containing $x^{2} + y^{2}$ .

∭_{V} f (x, y, z) d V

With $f = 1$ it gives volume; with density $ρ$ it gives mass.

Coordinates	Transformation	Volume element
Cartesian	—	$d x d y d z$
Cylindrical	$x = r cos θ$ , $y = r sin θ$ , $z = z$	$r d r d θ d z$
Spherical	$x = r sin ϕ cos θ$ , $y = r sin ϕ sin θ$ , $z = r cos ϕ$	$r^{2} sin ϕ d r d ϕ d θ$

( $ϕ$ measured from the positive z-axis, $0 \leq ϕ \leq π$ ; $θ$ in the xy-plane, $0 \leq θ \leq 2 π$ .)

FormulaPhysical applications (plane lamina of density ρ; area when ρ = 1)

Area: $A = \iint_{R} d A$ ; volume: $V = ∭ d V$ or $\iint z d A$
Mass: $M = \iint ρ d A$
Centroid / centre of mass: $\overset{x}{ˉ} = \frac{\iint x ρ d A}{M}$ , $\overset{y}{ˉ} = \frac{\iint y ρ d A}{M}$
Moments of inertia: $I_{x} = \iint y^{2} ρ d A$ , $I_{y} = \iint x^{2} ρ d A$
Polar moment of inertia: $I_{z} = J = \iint (x^{2} + y^{2}) ρ d A = I_{x} + I_{y}$ (perpendicular axis theorem)

Using polar coordinates:

(\int_{- \infty}^{\infty} e^{- x^{2}} d x)^{2} = \iint_{R^{2}} e^{- (x^{2} + y^{2})} d x d y = \int_{0}^{2 π} \int_{0}^{\infty} e^{- r^{2}} r d r d θ = π

\Rightarrow \int_{- \infty}^{\infty} e^{- x^{2}} d x = π

— the basis of the normal distribution.

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