Double Integrals

Why nesting works: with $x$ fixed, the inner integral $\int f(x, y)\,dy$ is the area of the slice of the solid under $z = f$ standing above the line at $x$ — a cross-sectional area $A(x)$. The outer integral $\int A(x)\,dx$ then adds up the slices, the way a loaf's volume is the sum of its slices. Fubini's theorem says slicing the other way gives the same volume. For the triangle under $y = x$ on $[0, 1]$, the slice at $x$ runs from $y = 0$ up to $y = x$, and those are the inner limits.

Why the $r$: the circles $r =$ const and the rays $\theta =$ const cut the plane into cells whose size depends on how far out they are. A cell at radius $r$ spanning an angle $\Delta\theta$ is $r\,\Delta\theta$ wide along the arc and $\Delta r$ deep, so its area is about $r\,\Delta r\,\Delta\theta$: cells far from the origin are bigger, and the factor $r$ records it. For the disk of radius $3$, $\int_0^{2\pi}\!\int_0^3 r\,dr\,d\theta = 2\pi\cdot\tfrac92 = 9\pi$, the correct area; dropping the $r$ gives $2\pi\cdot3 = 6\pi$.

Why it factors: in the inner integral $\int_0^3 xy\,dy$, $x$ is a constant and comes out, leaving $x\cdot\tfrac92$. The outer integral is then $\tfrac92\int_0^2 x\,dx = \tfrac92\cdot2 = 9$. This shortcut needs both conditions: an integrand of the form $g(x)h(y)$ and constant limits. Over the triangle below $y = x$, the same $xy$ does not split, because the inner limit $x$ ties the two integrals together.

Why redraw: the limits encode the shape, and describing a shape column by column uses different formulas from describing it row by row. The inner limits are the curves a line crosses as it passes through the region, so they change when the lines change direction; the outer limits are the extreme positions of those lines. Swapping the symbols without looking at the region almost never produces the same region.

That is $\tfrac12\left(\tfrac{32}{3} - \tfrac{32}{5}\right) = \tfrac12\cdot\tfrac{64}{15} = \tfrac{32}{15}$. In rows: for $0 \le y \le 4$, a horizontal line enters at the line $x = \tfrac y2$ and leaves at the parabola $x = \sqrt y$, so the integral is $\displaystyle\int_0^4 y\left(\sqrt y - \tfrac y2\right)dy = \int_0^4\left(y^{3/2} - \tfrac{y^2}{2}\right)dy = \tfrac{64}{5} - \tfrac{32}{3} = \tfrac{32}{15}$. Note that the curve that was the lower limit in columns is the right-hand limit in rows.

In columns, the line at $x$ runs from $y = 0$ to $y = x^2$, so the integral is $\displaystyle\int_0^1\!\int_0^{x^2}\sqrt{x^3 + 1}\,dy\,dx = \int_0^1 x^2\sqrt{x^3 + 1}\,dx$. The inner integration produced the factor $x^2$, which is exactly what the substitution $u = x^3 + 1$ needs: the integral is $\tfrac13\int_1^2\sqrt u\,du = \tfrac29\left(2^{3/2} - 1\right) = \tfrac29\left(2\sqrt2 - 1\right) \approx 0.406$.

Why this handles off-centre shapes: a boundary given as $r = r(\theta)$ is crossed once by each ray, so rays play the part vertical lines played before. The circle $x^2 + y^2 = 2x$, for instance, becomes $r^2 = 2r\cos\theta$, that is $r = 2\cos\theta$: a circle through the origin, traced once as $\theta$ runs from $-\tfrac\pi2$ to $\tfrac\pi2$.

So the integral is $\tfrac83\int_{-\pi/2}^{\pi/2}\cos^4\theta\,d\theta = \tfrac83\cdot\tfrac{3\pi}{8} = \pi$. Check: the disk has area $\pi$ and centre $(1, 0)$, so the average value of $x$ over it is $1$, and $\iint_D x\,dA = \pi\cdot1 = \pi$. The area itself comes out the same way: $\int_{-\pi/2}^{\pi/2}\!\int_0^{2\cos\theta} r\,dr\,d\theta = \int_{-\pi/2}^{\pi/2} 2\cos^2\theta\,d\theta = \pi$.

All four limits are constants, so the integral splits: $\left(\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta\right)\left(\int_1^2 r^3\,dr\right) = \tfrac12\cdot\tfrac{16 - 1}{4} = \tfrac{15}{8}$. Rings and sectors are the rectangles of polar coordinates.

So its area is $\approx r\,\Delta r\,\Delta\theta$ (exactly $\tfrac12\left[(r+\Delta r)^2 - r^2\right]\Delta\theta = r\,\Delta r\,\Delta\theta + \tfrac12\Delta r^2\Delta\theta$, and the second term vanishes in the limit). Summing cells gives $dA = r\,dr\,d\theta$. $\blacksquare$

Order 1 (inner $y$):

Step 1 — inner: $\int_0^x 2y\,dy = \left[y^2\right]_0^x = x^2$.

Step 2 — outer: $\int_0^1 x^2dx = \tfrac13$.

Order 2 (inner $x$):

Step 1 — inner: $\int_y^1 2y\,dx = 2y(1 - y)$ (the integrand is constant in $x$, so multiply by the interval length).

Step 2 — outer: $\int_0^1 (2y - 2y^2)dy = 1 - \tfrac23 = \tfrac13$. Both orders give $\tfrac13$ — Fubini in action, and a complete template: inner antiderivative, evaluate at the $x$-dependent limits, then an ordinary outer integral.

Step 1 — translate to polar: $x^2 + y^2 = r^2$ and $dA = r\,dr\,d\theta$, so the integrand becomes $r^2\cdot r = r^3$.

Step 2 — limits: the full disk is $0 \le r \le 2$, $0 \le \theta \le 2\pi$.

Step 3 — inner integral: $\int_0^2 r^3dr = \left[\tfrac{r^4}{4}\right]_0^2 = 4$.

Step 4 — outer: $\int_0^{2\pi}4\,d\theta = 8\pi$.

Step 5 — the cautionary rerun: forgetting the extra $r$ gives $\int_0^{2\pi}\!\int_0^2 r^2dr\,d\theta = \tfrac{16\pi}{3} \approx 16.8$ instead of $8\pi \approx 25.1$ — a wrong answer with no warning sign attached. Writing "$dA = r\,dr\,d\theta$" before computing anything is the habit that prevents it.