Exercises — Changing order of integration

Read the limits. Outer: $0\le x\le 2$ (constants — good). Inner: $0\le y\le x$. What this says: for each fixed $x$, a vertical strip runs from the wall $y=0$ up to the slant $y=x$. The region is the triangle bounded below by $y=0$, above by the line $y=x$, on the right by $x=2$. Corners: $(0,0),(2,0),(2,2)$. Check a point: $(1,0.5)$ — is $0\le 1\le 2$? yes. Is $0\le 0.5\le 1$? yes. Inside. ✓

Inner variable is $x$, so we integrate across first ⇒ a horizontal strip. Because $dxdy$ means "$x$ inner, $y$ outer", the outer limits ($0$ to $3$) must be pure numbers, and the inner limits $h_1(y),h_2(y)$ are curves (left wall to right wall of the strip). This is a Type II (horizontally simple) description — exactly the form defined in the Type I / Type II box above (see also Double Integrals over General Regions).

Sketch: $0\le x\le 1$, $x\le y\le 1$ — triangle above $y=x$, below $y=1$, right of $x=0$; corners $(0,0),(0,1),(1,1)$ (this is exactly the triangle in figure s01). New outer $y$: overall $0\le y\le 1$. New inner $x$: fix a height $y$. The horizontal strip enters at the wall $x=0$ and exits at the slant $y=x$. Solve the slant for the inner variable: $y=x\Rightarrow x=y$. So $0\le x\le y$. $\boxed{{\int }_0^1{\int }_0^{y}f(x,y)dxdy.}$ Check corner $(0.2,0.5)$: original needs $0.2\le 0.5\le 1$ ✓; new needs $0\le 0.2\le 0.5$ ✓. Same point, both descriptions accept it.

Sketch: $0\le x\le 2$, $0\le y\le x^2$ — under the parabola $y=x^2$, above $y=0$, left of $x=2$. Corners $(0,0),(2,0),(2,4)$. New outer $y$: the parabola reaches up to $y=x^2=2^2=4$, so $0\le y\le 4$. New inner $x$: fix $y$. Strip enters on the parabola and exits on the wall $x=2$. Solve $y=x^2$ for the inner variable, positive branch (region is $x\ge 0$): $x=\sqrt{y}$. So $\sqrt{y}\le x\le 2$. $\boxed{{\int }_0^4{\int }_{\sqrt{y}}^{2}fdxdy.}$

$\int e^{y^2}dy$ has no elementary antiderivative — this is a genuinely non-elementary integral in that order, so we must flip. Region (from L2.1's triangle): $0\le y\le 1$, $0\le x\le y$. $I={\int }_0^1{\int }_0^y{e}^{y^2}dxdy={\int }_0^1{e}^{y^2}[x{]}_0^{y}dy={\int }_0^{1}y{e}^{y^2}dy.$ Substitute $u=y^2,du=2ydy$: $I=\frac{1}{2}{\int }_0^1{e}^{u}du=\frac{1}{2}(e-1)$. $\boxed{I=\frac{e-1}{2}\approx \mathrm{0.8591.}}$ Why flipping worked: in the new order the strip length is exactly $y$, and $y{e}^{y^2}$ does have an antiderivative because the extra $y$ is the derivative-factor the substitution needs.

Same triangle again: $0\le y\le 1,0\le x\le y$. $K={\int }_0^1{\int }_0^y\frac{\sin y}{y}dxdy={\int }_0^1\frac{\sin y}{y}\cdot ydy={\int }_0^1\sin ydy=1-\cos 1.$ $\boxed{K=1-\cos 1\approx \mathrm{0.4597.}}$ The strip length $y$ cancels the $1/y$ — that cancellation is the entire reason to reorder.

Sketch: between parabola $y=x^2$ (below) and line $y=x$ (above); they meet at $(0,0)$ and $(1,1)$ (a lens — see figure s03). Why $x^2\le x$ here: on $0\le x\le 1$, squaring shrinks, so $x^2\le x$ — the parabola sits under the line. ✓ New outer $y$: $0\le y\le 1$. New inner $x$: fix $y$. Strip enters on the line $y=x\Rightarrow x=y$ (left edge) and exits on the parabola $y=x^2\Rightarrow x=\sqrt{y}$ (right edge). For $0<y<1$, $y<\sqrt{y}$, so $y\le x\le \sqrt{y}$. $\boxed{{\int }_0^1{\int }_y^{\sqrt{y}}fdxdy.}$

As given (Type II, horizontal strips): the overall $y$-range is $0\le y\le 2$. Fix $y$; the strip enters on the slant $x=y$ and exits on the wall $x=2$. One clean strip, no splitting: ${\int }_0^2{\int }_y^{2}fdxdy.$ Flip to Type I (vertical strips): the overall $x$-range is $0\le x\le 2$. Fix $x$; the strip enters on the wall $y=0$ and exits on the slant $y=x$ (solve $x=y$ for the inner variable ⇒ $y=x$). One clean strip again: ${\int }_0^2{\int }_0^{x}fdydx.$ Because $R$ is a single triangle, neither order needs splitting; only the inner limits differ (the slant $x=y$ acts as a ceiling in one order, a floor in the other). Same tiles, different bookkeeping.

The inner $\int x^3{e}^{x^5}dx$ resists (you'd want $x^4$ next to $e^{x^5}$, not $x^3$) — so flip. Region: $0\le y\le 4$, $\sqrt{y}\le x\le 2$ ⇒ this is L2.2's region ($0\le x\le 2$, $0\le y\le x^2$). $L={\int }_0^2{\int }_0^{x^2}{x}^3{e}^{x^5}dydx={\int }_0^2{x}^3{e}^{x^5}[y{]}_0^{x^2}dx={\int }_0^2{x}^5{e}^{x^5}dx.$ The strip length is $x^2$, giving $x^3\cdot x^2=x^5$. A clean substitution wants $u=x^5,du=5x^{4}dx$ — but we have $x^5$, not $x^4$, so this integrand does not close in elementary terms. The reorder is legal but not sufficient; we finish numerically. $\boxed{L={\int }_0^2{x}^5{e}^{x^5}dx\approx 8.09\times 10^{12}.}$ (The value is enormous because $e^{x^5}$ at $x=2$ is $e^{32}\approx 7.9\times 10^{13}$.)

Read boundaries: for fixed $y\in [0,1]$, $x$ runs from the line $x=y$ out to $x=\sqrt{2-y^2}$, i.e. $x^2+y^2=2$ — a circle of radius $\sqrt{2}$. Region: inside the circle $r=\sqrt{2}$, between the line $y=x$ and the $x$-axis. The line $x=y$ meets the circle where $2y^2=2\Rightarrow y=1$, point $(1,1)$. At $y=0$: $x$ from $0$ to $\sqrt{2}$.

Step A — the Cartesian $dx\leftrightarrow dy$ swap (what the title asks). Sketch the region: it is bounded on the left by the line $x=y$, on the right/top by the circular arc $x^2+y^2=2$, and below by $y=0$. To flip to $dydx$ (vertical strips), split by where the top of the strip changes curve:

• For $0\le x\le 1$: a vertical strip enters at $y=0$ and exits at the line $y=x$ (below the line is the region). So $0\le y\le x$.
• For $1\le x\le \sqrt{2}$: the line has passed the corner $(1,1)$; the strip now exits on the circle $y=\sqrt{2-x^2}$. So $0\le y\le \sqrt{2-x^2}$. $M={\int }_0^1{\int }_0^x\frac{dydx}{x^2+y^2}+{\int }_1^{\sqrt{2}}{\int }_0^{\sqrt{2-x^2}}\frac{dydx}{x^2+y^2}.$ Notice the Cartesian swap forces a split into two pieces — a sign the region wants a different coordinate system.

Step B — switch to polar (the clean route). In polar coordinates a point is $(x,y)=(r\cos \theta ,r\sin \theta )$, so the line $y=x$ means $r\sin \theta =r\cos \theta$, i.e. $\tan \theta =1$, whose first-quadrant solution is $\theta =\pi /4$. A line through the origin is exactly "one fixed angle", which is why it becomes a constant $\theta$-limit. So $R$ is the sector $0\le \theta \le \pi /4$, $0\le r\le \sqrt{2}$ — see Polar Coordinates in Double Integrals. The integrand $\frac{1}{x^2+y^2}=\frac{1}{r^2}$ and $dA=rdrd\theta$: $M={\int }_0^{\pi /4}{\int }_0^{\sqrt{2}}\frac{1}{r^2}rdrd\theta ={\int }_0^{\pi /4}{\int }_0^{\sqrt{2}}\frac{1}{r}drd\theta .$ But ${\int }_0^{\sqrt{2}}\frac{1}{r}dr=[\ln r{]}_0^{\sqrt{2}}$ diverges at $r=0$ — the integrand blows up at the origin, a corner of $R$. So $\boxed{M\text{ diverges }(=+\infty ).}$ Mastery point: the Cartesian swap already warned us (it split into pieces); polar (via the Jacobian) then exposed the true difficulty — a singularity at the origin — that both Cartesian forms hid. Changing order/coordinates is a diagnostic tool, not just a computation trick.

Region: $0\le x\le 1$, $0\le y\le 1-x$ ⇒ triangle under the line $x+y=1$; corners $(0,0),(1,0),(0,1)$. Flip: new outer $y\in [0,1]$; fix $y$, strip from $x=0$ to $x=1-y$ (solve $x+y=1$ for $x$). $\boxed{{\int }_0^1{\int }_0^{1-y}fdxdy.}$ 3D generalisation: a triple integral over a solid can be summed in any of the $3!=6$ orders of $dx,dy,dz$; each order re-describes the same solid by choosing which variable's overall range is the outermost constant and expressing the rest as functions of the outer variables — identical logic, one more layer.

Every problem: sketch $R$, choose the outer variable (its range = outer constants), fix it and read where the strip enters/exits (those curves, re-solved for the inner variable, = inner limits) — and stay alert for non-elementary inners and singularities.