Regions and order of integration Describe non-rectangular regions by slices and set up double integrals with the correct bounds.
This page extends double integrals from rectangles to regions bounded by
curves and shows how to read the limits of integration from a sketch.
The idea is to place a bounded region inside a rectangle and extend the
function by zero outside the region. When that zero extension is bounded and
piecewise continuous, with discontinuities only along boundary curves,
Fubini's theorem
applies to the enclosing rectangle because those curves have area zero.
Finding the limits is a separate geometric task: each slice must start and
stop on the boundary of the original region. When neither vertical nor
horizontal slices give workable bounds, reshape the region with a change of
variables .
[ DEFINITION ] Let Ω \Omega Ω R R R
f Ω ( x , y ) = { f ( x , y ) , ( x , y ) ∈ Ω , 0 , ( x , y ) ∈ R ∖ Ω . f_\Omega(x,y)=
\begin{cases}
f(x,y),&(x,y)\in\Omega,\\
0,&(x,y)\in R\setminus\Omega.
\end{cases} f Ω ( x , y ) = { f ( x , y ) , 0 , ( x , y ) ∈ Ω , ( x , y ) ∈ R ∖ Ω. If f Ω f_\Omega f Ω R R R Ω \Omega Ω
∬ Ω f ( x , y ) d A : = ∬ R f Ω ( x , y ) d A . \iint_\Omega f(x,y)\,dA
:=\iint_R f_\Omega(x,y)\,dA. ∬ Ω f ( x , y ) d A := ∬ R f Ω ( x , y ) d A . The zero extension contributes nothing outside Ω \Omega Ω Ω \Omega Ω
[RIGHT CLICK] [ASK AI] [ THEOREM ] If c ( x ) ≤ d ( x ) c(x)\le d(x) c ( x ) ≤ d ( x ) [ a , b ] [a,b] [ a , b ]
Ω = { ( x , y ) : a ≤ x ≤ b , c ( x ) ≤ y ≤ d ( x ) } , \Omega=\{(x,y):a\le x\le b,\ c(x)\le y\le d(x)\}, Ω = {( x , y ) : a ≤ x ≤ b , c ( x ) ≤ y ≤ d ( x )} , then
∬ Ω f ( x , y ) d A = ∫ a b ∫ c ( x ) d ( x ) f ( x , y ) d y d x . \iint_\Omega f(x,y)\,dA
=\int_a^b\int_{c(x)}^{d(x)}
f(x,y)\,dy\,dx. ∬ Ω f ( x , y ) d A = ∫ a b ∫ c ( x ) d ( x ) f ( x , y ) d y d x . For each fixed x x x y y y c ( x ) c(x) c ( x ) d ( x ) d(x) d ( x )
[RIGHT CLICK] [ASK AI] [ THEOREM ] If a ( y ) ≤ b ( y ) a(y)\le b(y) a ( y ) ≤ b ( y ) [ c , d ] [c,d] [ c , d ]
Ω = { ( x , y ) : c ≤ y ≤ d , a ( y ) ≤ x ≤ b ( y ) } , \Omega=\{(x,y):c\le y\le d,\ a(y)\le x\le b(y)\}, Ω = {( x , y ) : c ≤ y ≤ d , a ( y ) ≤ x ≤ b ( y )} , then
∬ Ω f ( x , y ) d A = ∫ c d ∫ a ( y ) b ( y ) f ( x , y ) d x d y . \iint_\Omega f(x,y)\,dA
=\int_c^d\int_{a(y)}^{b(y)}
f(x,y)\,dx\,dy. ∬ Ω f ( x , y ) d A = ∫ c d ∫ a ( y ) b ( y ) f ( x , y ) d x d y . For each fixed y y y x x x a ( y ) a(y) a ( y ) b ( y ) b(y) b ( y )
[RIGHT CLICK] [ASK AI] [ METHOD ] To set up a double integral over Ω \Omega Ω
sketch the region and every boundary curve;
decide whether vertical or horizontal slices give simpler bounds;
find the range of the outside variable;
fix that variable and find the two inner bounds;
write the inner integral first, then integrate from the inside outward;
split the region if one pair of inner bounds does not describe every slice.
Changing the order describes the same region by different slices, so the new
limits must be read from the boundary, not by swapping the old ones.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Let
f ( x , y ) = x y 2 f(x,y)=xy^2 f ( x , y ) = x y 2 and let Ω \Omega Ω
( 0 , 0 ) , ( 1 , 1 ) , ( 1 , − 1 ) . (0,0),\qquad(1,1),\qquad(1,-1). ( 0 , 0 ) , ( 1 , 1 ) , ( 1 , − 1 ) . Vertical slices give
0 ≤ x ≤ 1 , − x ≤ y ≤ x . 0\le x\le1,\qquad -x\le y\le x. 0 ≤ x ≤ 1 , − x ≤ y ≤ x . Therefore
∬ Ω x y 2 d A = ∫ 0 1 ∫ − x x x y 2 d y d x = ∫ 0 1 x [ y 3 3 ] − x x d x = ∫ 0 1 2 x 4 3 d x = 2 15 . \begin{aligned}
\iint_\Omega xy^2\,dA
&=\int_0^1\int_{-x}^{x}xy^2\,dy\,dx\\
&=\int_0^1x\left[\frac{y^3}{3}\right]_{-x}^{x}dx\\
&=\int_0^1\frac{2x^4}{3}\,dx
=\frac{2}{15}.
\end{aligned} ∬ Ω x y 2 d A = ∫ 0 1 ∫ − x x x y 2 d y d x = ∫ 0 1 x [ 3 y 3 ] − x x d x = ∫ 0 1 3 2 x 4 d x = 15 2 . The same triangle can be described horizontally by
− 1 ≤ y ≤ 1 , ∣ y ∣ ≤ x ≤ 1. -1\le y\le1,\qquad |y|\le x\le1. − 1 ≤ y ≤ 1 , ∣ y ∣ ≤ x ≤ 1. The reversed-order calculation is
∬ Ω x y 2 d A = ∫ − 1 1 ∫ ∣ y ∣ 1 x y 2 d x d y = ∫ − 1 1 y 2 2 ( 1 − y 2 ) d y = 2 15 . \begin{aligned}
\iint_\Omega xy^2\,dA
&=\int_{-1}^1\int_{|y|}^{1}xy^2\,dx\,dy\\
&=\int_{-1}^1\frac{y^2}{2}(1-y^2)\,dy
=\frac{2}{15}.
\end{aligned} ∬ Ω x y 2 d A = ∫ − 1 1 ∫ ∣ y ∣ 1 x y 2 d x d y = ∫ − 1 1 2 y 2 ( 1 − y 2 ) d y = 15 2 . [RIGHT CLICK] [ASK AI] [ EXAMPLE ] Consider
∫ 0 1 ∫ 2 x 2 sin ( y 2 ) d y d x . \int_0^1\int_{2x}^{2}\sin(y^2)\,dy\,dx. ∫ 0 1 ∫ 2 x 2 sin ( y 2 ) d y d x . The inner integral has no elementary antiderivative. The region is
0 ≤ x ≤ 1 , 2 x ≤ y ≤ 2. 0\le x\le1,\qquad 2x\le y\le2. 0 ≤ x ≤ 1 , 2 x ≤ y ≤ 2. Its y y y 0 0 0 2 2 2 y y y 2 x ≤ y 2x\le y 2 x ≤ y 0 ≤ x ≤ y / 2 0\le x\le y/2 0 ≤ x ≤ y /2
∫ 0 1 ∫ 2 x 2 sin ( y 2 ) d y d x = ∫ 0 2 ∫ 0 y / 2 sin ( y 2 ) d x d y = ∫ 0 2 y 2 sin ( y 2 ) d y = 1 4 ∫ 0 4 sin u d u ( u = y 2 ) = 1 − cos 4 4 . \begin{aligned}
\int_0^1\int_{2x}^{2}\sin(y^2)\,dy\,dx
&=\int_0^2\int_0^{y/2}\sin(y^2)\,dx\,dy\\
&=\int_0^2\frac y2\sin(y^2)\,dy\\
&=\frac14\int_0^4\sin u\,du
\qquad(u=y^2)\\
&=\frac{1-\cos4}{4}.
\end{aligned} ∫ 0 1 ∫ 2 x 2 sin ( y 2 ) d y d x = ∫ 0 2 ∫ 0 y /2 sin ( y 2 ) d x d y = ∫ 0 2 2 y sin ( y 2 ) d y = 4 1 ∫ 0 4 sin u d u ( u = y 2 ) = 4 1 − cos 4 . Reversing the order turns the difficult inner integral into a direct substitution.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Consider
∫ − 1 1 ∫ x 2 2 − x 2 f ( x , y ) d y d x . \int_{-1}^{1}\int_{x^2}^{2-x^2}f(x,y)\,dy\,dx. ∫ − 1 1 ∫ x 2 2 − x 2 f ( x , y ) d y d x . The region lies between
y = x 2 , y = 2 − x 2 . y=x^2,\qquad y=2-x^2. y = x 2 , y = 2 − x 2 . Its y y y 0 ≤ y ≤ 2 0\le y\le2 0 ≤ y ≤ 2 0 ≤ y ≤ 1 0\le y\le1 0 ≤ y ≤ 1 x 2 ≤ y x^2\le y x 2 ≤ y
− y ≤ x ≤ y . -\sqrt y\le x\le\sqrt y. − y ≤ x ≤ y . For 1 ≤ y ≤ 2 1\le y\le2 1 ≤ y ≤ 2 y ≤ 2 − x 2 y\le2-x^2 y ≤ 2 − x 2
− 2 − y ≤ x ≤ 2 − y . -\sqrt{2-y}\le x\le\sqrt{2-y}. − 2 − y ≤ x ≤ 2 − y . The reversed order is therefore
∫ − 1 1 ∫ x 2 2 − x 2 f ( x , y ) d y d x = ∫ 0 1 ∫ − y y f ( x , y ) d x d y + ∫ 1 2 ∫ − 2 − y 2 − y f ( x , y ) d x d y . \begin{aligned}
\int_{-1}^{1}\int_{x^2}^{2-x^2}f(x,y)\,dy\,dx
&=\int_0^1\int_{-\sqrt y}^{\sqrt y}f(x,y)\,dx\,dy\\
&\quad+\int_1^2\int_{-\sqrt{2-y}}^{\sqrt{2-y}}
f(x,y)\,dx\,dy.
\end{aligned} ∫ − 1 1 ∫ x 2 2 − x 2 f ( x , y ) d y d x = ∫ 0 1 ∫ − y y f ( x , y ) d x d y + ∫ 1 2 ∫ − 2 − y 2 − y f ( x , y ) d x d y . A single pair of horizontal bounds would either omit part of the region or
include points outside it.
[RIGHT CLICK] [ASK AI] [ PROPOSITION ] For integrable f f f g g g Ω \Omega Ω k k k
∬ Ω ( f + g ) d A = ∬ Ω f d A + ∬ Ω g d A , \iint_\Omega(f+g)\,dA
=\iint_\Omega f\,dA+\iint_\Omega g\,dA, ∬ Ω ( f + g ) d A = ∬ Ω f d A + ∬ Ω g d A , ∬ Ω k f d A = k ∬ Ω f d A . \iint_\Omega kf\,dA
=k\iint_\Omega f\,dA. ∬ Ω k f d A = k ∬ Ω f d A . If Ω = Ω 1 ∪ Ω 2 \Omega=\Omega_1\cup\Omega_2 Ω = Ω 1 ∪ Ω 2
∬ Ω f d A = ∬ Ω 1 f d A + ∬ Ω 2 f d A . \iint_\Omega f\,dA
=\iint_{\Omega_1}f\,dA+\iint_{\Omega_2}f\,dA. ∬ Ω f d A = ∬ Ω 1 f d A + ∬ Ω 2 f d A . Splitting the parabolic region above is an instance of additivity.
[RIGHT CLICK] [ASK AI] [ COROLLARY ] The area of a planar region Ω \Omega Ω
Area ( Ω ) = ∬ Ω 1 d A . \operatorname{Area}(\Omega)=\iint_\Omega 1\,dA. Area ( Ω ) = ∬ Ω 1 d A . The integrand 1 1 1 1 1 1 mass
instead of area.
[RIGHT CLICK] [ASK AI] Regions and order of integration Builds on Double integrals and Riemann sums .