Multiple integrals · Definition
Fubini's theorem
Before you start you need to know
It turns a double integral, impractical to compute with sums, into two single-variable integrals you already know how to solve.
Definition
If \(f\) is continuous on a rectangle \(R=[a,b]\times[c,d]\), its double integral with respect to the area \(A\) is computed as an iterated integral, integrating in either order.
Fixing \(x\) and computing \(A(x)=\int_c^d f(x,y)\,dy\) measures the area of the slice of the solid above the vertical line at that \(x\). Integrating \(A(x)\) from \(a\) to \(b\) adds up all the slices and gives the volume. This is Cavalieri's principle, and it is what the theorem says: the volume does not depend on whether you slice parallel to one axis or the other.
In practice you integrate from the inside out: in the inner integral, the other variable is treated as a constant. Both orders give the same number, but one can be much simpler; for \(x e^{xy}\), for instance, it pays to integrate in \(y\) first, since an antiderivative in \(x\) would require integration by parts.
When \(f(x,y)=g(x)h(y)\) on a rectangle, the double integral factors: \(\iint_R f\,dA=\int_a^b g\,dx\cdot\int_c^d h\,dy\). Guiding example: \(\iint_R xy\,dA\) with \(R=[0,2]\times[0,3]\) equals \(2\cdot\tfrac92=9\). This factoring is valid only on rectangles and for functions that are a product of a function of \(x\) and a function of \(y\).
Typical mistakes: factoring when \(f\) is not a product (for instance \(x+y\)), forgetting that each integral's limits belong to its own variable, and applying the theorem without continuity. Regions that are not rectangles are handled in the double integral over a general region.
Think about it
Why is \(\iint_R(x+y)\,dA\) not \(\int_a^b(x+y)\,dx\cdot\int_c^d(x+y)\,dy\)?
Because factoring requires \(f\) to be a product \(g(x)h(y)\); \(x+y\) is a sum. With Fubini you compute \(\int_a^b\!\int_c^d(x+y)\,dy\,dx\) directly.
What is the advantage of choosing the order of integration?
One order may give an elementary antiderivative while the other forces integration by parts or has no elementary antiderivative; the final result is the same.
Exercises 10
- Compute \(\iint_R xy\,dA\) over \(R=[0,2]\times[0,3]\).
- Compute \(\iint_R (x+y)\,dA\) over \(R=[0,1]\times[0,2]\).
- Compute \(\iint_R e^{x+y}\,dA\) over \(R=[0,1]\times[0,1]\).
- Compute \(\iint_R x^2y\,dA\) over \(R=[1,2]\times[0,1]\).
- Compute \(\iint_R \sin x\cos y\,dA\) over \(R=[0,\tfrac{\pi}{2}]\times[0,\tfrac{\pi}{2}]\).
- Compute \(\iint_R 6xy^2\,dA\) over \(R=[0,1]\times[-1,1]\).
- Let \(f\) be continuous on \(R=[a,b]\times[c,d]\). What does Fubini's theorem state?
- Compute \(\iint_R xe^{xy}\,dA\) over \(R=[0,1]\times[0,1]\). It helps to integrate in \(y\) first.
- If \(f(x,y)=g(x)h(y)\) with \(g,h\) continuous, what is \(\iint_R f\,dA\) over \(R=[a,b]\times[c,d]\)?
- Compute \(\iint_R y\cos(xy)\,dA\) over \(R=[0,1]\times[0,\pi]\). Choose the order wisely.
What comes next
References
- James Stewart, Daniel K. Clegg, Saleem Watson, Calculus: Early Transcendentals, 9th edition, Cengage, 2021. §15.2
- 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. §5.2