Multiple integrals · Definition

Double integral over a general region

multivariable calculusmultiple integrals

Before you start you need to know

It lets you integrate over regions with curved boundaries, not just rectangles, by describing them as lying «between two curves».

Definition

The double integral of \(f\) with respect to the area \(A\) over a region \(D=\{a\le x\le b,\ g_1(x)\le y\le g_2(x)\}\) is an iterated integral whose inner limits depend on the outer variable.

\[\iint_D f\,dA=\int_a^b\!\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx\]

A type I region lies between two graphs \(y=g_1(x)\) and \(y=g_2(x)\) with \(a\le x\le b\); a type II region lies between \(x=h_1(y)\) and \(x=h_2(y)\) with \(c\le y\le d\), and is integrated as \(\int_c^d\!\int_{h_1(y)}^{h_2(y)} f\,dx\,dy\). The idea is the same as on a rectangle: for each value of the outer variable, the inner one sweeps a segment of the region, and that segment's endpoints are the inner limits.

Rule of thumb: outer limits are numbers, inner limits may depend on the outer variable, and the final result is a number. To find them, sketch the region, draw a typical segment in the inner integration direction and read where it enters and exits. Guiding example: the area between \(y=x^2\) and \(y=2x\) is \(\int_0^2\!\int_{x^2}^{2x}dy\,dx=\int_0^2(2x-x^2)\,dx=\tfrac43\).

If the region is of both types you can choose the order, and switching sometimes turns an impossible integral into an easy one: \(\int_0^1\!\int_x^1e^{y^2}\,dy\,dx\) has no elementary antiderivative in \(y\), but describing the region as \(0\le x\le y\le1\) turns it into \(\int_0^1ye^{y^2}\,dy=\tfrac{e-1}{2}\). With \(f=1\) you get the area of \(D\).

Typical mistakes: putting variable limits on the outer integral, not sketching the region and mixing up which curve is the upper one, and, when switching order, forgetting to rewrite the limits: merely swapping \(dx\) and \(dy\) is not enough.

Think about it

Why can't the outer integral have limits that depend on the inner variable?

Because integrating over the outer variable must produce a number; if its limits depended on the already-integrated inner variable, the expression would make no sense.

Describe \(D=\{0\le x\le1,\ x^2\le y\le x\}\) as a type II region.

For each \(y\in[0,1]\), \(x\) runs from \(y\) to \(\sqrt{y}\): \(D=\{0\le y\le1,\ y\le x\le\sqrt y\}\).

Exercises 10

  1. Compute \(\iint_D (x+y)\,dA\), where \(D=\{0\le x\le1,\ 0\le y\le x\}\).
  2. Compute \(\iint_D 2xy\,dA\), where \(D=\{0\le x\le1,\ x^2\le y\le x\}\).
  3. Find the area of the region between \(y=x^2\) and \(y=2x\), that is, \(\iint_D dA\).
  4. Compute \(\iint_D x\,dA\), where \(D=\{0\le x\le2,\ 0\le y\le x^2\}\).
  5. The region \(D=\{0\le y\le x\le1\}\) can also be described by integrating in \(x\) first. Which iterated integral equals \(\int_0^1\!\int_0^x f\,dy\,dx\)?
  6. Compute \(\int_0^1\!\int_y^1 \sin(x^2)\,dx\,dy\) by reversing the order of integration.
  7. Compute \(\iint_D xy\,dA\), where \(D=\{0\le y\le1,\ y\le x\le2\}\).
  8. Find the area of \(D=\{1\le x\le e,\ 0\le y\le\ln x\}\).
  9. Let \(D\) be the region between the graphs \(y=g_1(x)\) and \(y=g_2(x)\), with \(g_1\le g_2\) and \(a\le x\le b\). Which is the correct iterated integral of \(f\) over \(D\)?
  10. Compute \(\int_0^1\!\int_x^1 e^{y^2}\,dy\,dx\). Reverse the order of integration.

What comes next

References

  • James Stewart, Daniel K. Clegg, Saleem Watson, Calculus: Early Transcendentals, 9th edition, Cengage, 2021. §15.3
  • Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr., Thomas' Calculus, 15th edition, Pearson, 2023. §15.2
  • Jerrold E. Marsden, Anthony Tromba, Vector Calculus, 6th edition, W. H. Freeman, 2012. §5.3