Worked examples — Double integrals over rectangles — Fubini's theorem

This page is a drill of cases. The parent note built the machine (freeze → fill → file). Here we run that machine through every kind of input a rectangle can throw at you: positive surfaces, surfaces that dip below the plane, zero answers, degenerate rectangles, separable shortcuts, a physics word problem, and an exam twist where the order of integration decides whether you finish in one line or one hour.

Before we start, one reminder in plain words. A double integral ${\iint }_{R}fdA$ is the signed volume trapped between the surface $z=f(x,y)$ and the flat floor (the $xy$-plane) over the rectangle $R$. "Signed" means: where the surface is above the floor the volume counts positive, where it dips below the floor it counts negative. Keep that picture — most of the cases below are just which sign wins.

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The scenario matrix

Every rectangular double integral you meet falls into one of these cells. The worked examples are tagged with the cell they cover.

#	Case class	What's tricky about it	Example
A	All-positive integrand	Surface entirely above floor → answer is a genuine positive volume	Ex 1
B	Sign-changing integrand	Surface crosses the floor → signed cancellation, answer can be small or zero	Ex 2
C	Answer is exactly zero (symmetry)	Positive lump exactly cancels negative lump	Ex 3
D	Separable shortcut	$f=g(x)h(y)$ → split into two 1-D integrals	Ex 4
E	Degenerate rectangle ($a=b$ or $c=d$)	Zero-width slab → volume must be $0$	Ex 5
F	Order matters for effort	One order is one line, the other needs parts	Ex 6
G	Real-world word problem (units!)	Translate a physical quantity into $\iint$; carry units	Ex 7
H	Exam twist: constant / average value	Reinterpret the integral as an average height	Ex 8

We will hit all eight cells.

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Ex 1 — Cell A: all-positive integrand (a real volume)

Look at the surface first — it is a valley floor curving upward in both directions.

Step 1 — Freeze $x$, integrate over $y$ (fill the inner integral). ${\int }_0^1(x^2+y^2)dy={\left[x^{2}y+\frac{y^3}{3}\right]}_{y=0}^{y=1}=x^2+\frac{1}{3}.$ Why this step? Freezing $x$ turns the inner job into an ordinary 1-D integral in $y$; the term $x^2$ is a frozen constant, so its antiderivative in $y$ is $x^{2}y$. The result $A(x)=x^2+\frac{1}{3}$ is the cross-sectional area of the solid at position $x$.

Step 2 — File $A(x)$ into the outer integral. ${\int }_0^1\left(x^2+\frac{1}{3}\right)dx={\left[\frac{x^3}{3}+\frac{x}{3}\right]}_0^1=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}.$ Why this step? Adding up all the thin slabs $A(x)dx$ is exactly Volume by slicing (single-variable) — one dimension down.

Verify: Swap the order. By symmetry ($x\leftrightarrow y$ leaves the integrand unchanged) the inner-$x$ integral also gives $y^2+\frac{1}{3}$, and the outer gives $\frac{2}{3}$ again. Answer $\frac{2}{3}\approx 0.667$ sits in the predicted range $[0,2]$. ✅

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Ex 2 — Cell B: sign-changing integrand

The red patch (below floor, $x<y$) is a small triangle near the origin; the blue region (above floor) dominates.

Step 1 — Inner integral over $y$ (freeze $x$). ${\int }_0^1(x-y)dy={\left[xy-\frac{y^2}{2}\right]}_0^1=x-\frac{1}{2}.$ Why this step? Constant $x$ integrates to $xy$; the genuinely $y$-dependent term $-y$ integrates to $-\frac{y^2}{2}$. Note $A(x)=x-\frac{1}{2}$ is itself negative for $x<\frac{1}{2}$ — that is the below-floor part showing up.

Step 2 — Outer integral over $x$. ${\int }_0^2\left(x-\frac{1}{2}\right)dx={\left[\frac{x^2}{2}-\frac{x}{2}\right]}_0^2=(2-1)-0=1.$

Verify: Split into signed pieces. $\iint xdA=\left({\int }_0^{2}xdx\right)\left({\int }_0^{1}dy\right)=2\cdot 1=2$ and $\iint ydA=\left({\int }_0^{2}dx\right)\left({\int }_0^{1}ydy\right)=2\cdot \frac{1}{2}=1$. Difference $2-1=1$. ✅ Positive and small, as forecast.

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Ex 3 — Cell C: answer is exactly zero (symmetry)

Step 1 — Recognise it's separable, split it. ${\iint }_R\sin x\cos ydA=\left({\int }_{-\pi }^{\pi }\sin xdx\right)\left({\int }_0^{\pi }\cos ydy\right).$ Why this step? $f=g(x)h(y)$ with $g=\sin x$, $h=\cos y$; the product rule for separable integrands (parent note) turns one 2-D job into two 1-D jobs.

Step 2 — Evaluate the $x$-factor. ${\int }_{-\pi }^{\pi }\sin xdx=[-\cos x{]}_{-\pi }^{\pi }=-\cos \pi +\cos (-\pi )=-(-1)+(-1)=0.$ Why this step? Any factor being zero forces the whole product to zero — no need to finish the $y$-integral. This is the odd-function-over-symmetric-interval cancellation made rigorous.

Verify: The $y$-factor is ${\int }_0^{\pi }\cos ydy=[\sin y{]}_0^{\pi }=0-0=0$ too — but we only needed one zero factor. Product $=0\times 0=0$. ✅

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Ex 4 — Cell D: separable shortcut with a twist

Step 1 — Split into a product of two 1-D integrals. ${\iint }_{R}x{e}^{x}ydA=\left({\int }_0^{1}x{e}^{x}dx\right)\left({\int }_0^{2}ydy\right).$ Why this step? Variables never mix and both limits are constants, so the double integral factors — the whole point of Cell D.

Step 2 — The $x$-factor by parts. Choose $u=x,dv=e^{x}dx$, so $du=dx,v=e^x$: ${\int }_0^{1}x{e}^{x}dx=[x{e}^x{]}_0^1-{\int }_0^1{e}^{x}dx=e-(e-1)=1.$ Why this step? Integration by parts is the tool for a product "polynomial $\times$ exponential" — differentiating $x$ kills the polynomial factor in one step, which is exactly what parts is designed to exploit.

Step 3 — The $y$-factor and the product. ${\int }_0^{2}ydy={\left[\frac{y^2}{2}\right]}_0^2=2,\text{answer}=1\times 2=2.$

Verify: Numerically ${\int }_0^{1}x{e}^{x}dx=e\cdot 1-(e-1)=1$ exactly; times $2$ gives $\boxed{2}$. ✅ Matches the forecast.

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Ex 5 — Cell E: degenerate rectangle (zero width)

Step 1 — Read the outer limits. ${\int }_2^2\left({\int }_0^5(x^2+3xy+7)dy\right)dx.$ Why this step? The inner integral produces some function $A(x)$ — it doesn't matter what — because the outer integral runs from $2$ to $2$.

Step 2 — Apply the zero-length rule. ${\int }_2^{2}A(x)dx=0\text{for any integrable }A.$ Why this step? ${\int }_a^{a}g=0$ is a fundamental property of definite integrals: no interval, no accumulation. Geometrically the "solid" is a flat sheet of zero thickness — volume $0$.

Verify: Even brute force agrees. Inner: ${\int }_0^5(x^2+3xy+7)dy=5x^2+\frac{75}{2}x+35$, a finite $A(x)$. Outer ${\int }_2^{2}A(x)dx=0$. ✅ This is the degenerate limiting case — always check for it before grinding algebra.

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Ex 6 — Cell F: order matters for effort

Step 1 — Integrate over $y$ first (freeze $x$). ${\int }_0^{1}x{e}^{xy}dy=[e^{xy}{]}_{y=0}^{y=1}=e^x-1.$ Why this step? Because ${\partial }_y{e}^{xy}=x{e}^{xy}$ exactly matches the integrand — the $x$ out front is precisely the chain-rule factor. Choosing $dy$ first is what makes this a one-liner. This is Fubini in action: same solid, cheaper slicing direction.

Step 2 — Integrate the result over $x$. ${\int }_0^1(e^x-1)dx=[e^x-x{]}_0^1=(e-1)-(1-0)=e-2.$

Verify: $e-2\approx 0.718$ — positive, below $1$, as forecast. Sanity-check the other order: ${\int }_0^{1}x{e}^{xy}dx$ needs integration by parts and produces a messier expression that also integrates to $e-2$ — same answer, far more work. ✅ Fubini earns its keep here.

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Ex 7 — Cell G: real-world word problem (carry units!)

Step 1 — Set up the mass integral. $M={\iint }_R\rho (x,y)dA={\int }_0^3{\int }_0^2(2+x+y)dydx[{\text{kg/m}}^2\cdot {\text{m}}^2=\text{kg}].$ Why this step? Each tiny tile of area $dA$ (m²) carries mass $\rho dA$ (kg). Summing tiles is precisely a 2-D Riemann sum whose limit is the double integral — the physical meaning of $\iint$.

Step 2 — Inner integral over $y$ (freeze $x$). ${\int }_0^2(2+x+y)dy=[(2+x)y+\frac{y^2}{2}{]}_0^2=2(2+x)+2=2x+6.$ Why this step? $2+x$ is a frozen constant in $y$, integrating to $(2+x)y$; the true $y$-term gives $\frac{y^2}{2}$.

Step 3 — Outer integral over $x$. $M={\int }_0^3(2x+6)dx=[x^2+6x{]}_0^3=9+18=27\text{kg}.$

Verify: Units: ${\text{kg/m}}^2\times {\text{m}}^2=\text{kg}$ ✅. Value $27\text{kg}$ matches the average-density estimate exactly (because $\rho$ is linear, the mean density really is the value at the centre $(1.5,1)$: $2+1.5+1=4.5$, times area $6$ = $27$). ✅

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Ex 8 — Cell H: exam twist (average value / constant height)

Step 1 — Recall the definition of average value. $\bar{f}=\frac{1}{\text{area}(R)}{\iint }_{R}fdA.$ Why this step? Average height = (total volume)$÷$(base area). This is the 2-D twin of the single-variable mean $\frac{1}{b-a}{\int }_a^{b}f$. It answers: "what flat height gives the same volume?"

Step 2 — Compute the volume (separable). ${\iint }_R{x}^{2}ydA=\left({\int }_0^2{x}^{2}dx\right)\left({\int }_0^{3}ydy\right)=\frac{8}{3}\cdot \frac{9}{2}=12.$ Why this step? $x^{2}y$ is separable, so it splits into two 1-D integrals — the fastest route.

Step 3 — Divide by the area. $\bar{f}=\frac{12}{2\cdot 3}=\frac{12}{6}=2.$

Verify: A box of base area $6$ and height $\bar{f}=2$ has volume $6\times 2=12$ — matching the computed volume. ✅ And $\bar{f}=2$ lies inside the range $[0,12]$, as forecast.

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Recall Quick self-test across all cells

Which cell has a guaranteed zero answer before computing? ::: Cell E (degenerate rectangle, $a=b$ or $c=d$) — and Cell C when an odd factor spans a symmetric interval. Mass of a plate needs which integral? ::: ${\iint }_R\rho dA$ (density times area, summed) — Cell G. When does choosing the integration order save real work? ::: Cell F — when one order's integrand is already a recognisable derivative (here ${\partial }_y{e}^{xy}=x{e}^{xy}$). Average value formula over $R$? ::: $\bar{f}=\frac{1}{\text{area}(R)}{\iint }_{R}fdA$.

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Connections

• Riemann sums — the word-problem mass (Ex 7) is a 2-D Riemann sum limit.
• Volume by slicing (single-variable) — every "file the inner into the outer" step (Ex 1).
• Change of order of integration — Cell F (Ex 6) chooses the cheaper slicing direction.
• Double integrals over general regions — next: non-constant inner limits (here they're always numbers).
• Fubini–Tonelli theorem — guarantees the order-swaps used above are legal.
• Triple integrals — add a third slicing direction for mass of a solid, not a plate.