Exercises — Area between curves — horizontal and vertical slices

WHAT we do: find where they meet, then test one interior point. Set them equal: $x+1=x^2-2x+1\Rightarrow 0=x^2-3x\Rightarrow x(x-3)=0$, so they meet at $x=0$ and $x=3$ (this uses Solving Quadratic & Polynomial Equations). Test a point at $x=1$ (interior): line gives $1+1=2$; parabola gives $1-2+1=0$. Since $2>0$, the line $y=x+1$ is on top across the whole interval $[0,3]$. Why one point is enough: the curves only cross at the two endpoints, so the order (who is on top) can’t flip anywhere in between. Answer: top $=y=x+1$, bottom $=y=x^2-2x+1$, meeting at $x=0,3$.

Look at the picture (s02): the parabola $x=y^2$ opens rightward. For a fixed height $y$, the region runs from the parabola on the left ($x=y^2$) to the line on the right ($x=4$). That is one right curve and one left curve for every $y$ — a clean horizontal slice. A vertical slice at fixed $x$ would hit the curve at $y=+\sqrt{x}$ and $y=-\sqrt{x}$ — top and bottom are two branches of the same curve, awkward to write as $f(x)$. $y$-limits: the parabola meets $x=4$ when $y^2=4$, i.e. $y=\pm 2$. So $c=-2,d=2$. Answer: horizontal slices; $y\in [-2,2]$, integrand $R(y)-L(y)=4-y^2$.

Intersections: $4-x^2=0\Rightarrow x^2=4\Rightarrow x=\pm 2$. So $a=-2,b=2$. Top/bottom: at $x=0$, curve $=4>0=$ axis, so the parabola is the top, $y=0$ the bottom. Integrate top minus bottom (vertical slices, thickness $dx$): $A={\int }_{-2}^2((4-x^2)-0)dx=[4x-\frac{x^3}{3}{]}_{-2}^2.$ At $x=2$: $8-\frac{8}{3}=\frac{16}{3}$. At $x=-2$: $-8+\frac{8}{3}=-\frac{16}{3}$. $A=\frac{16}{3}-(-\frac{16}{3})=\frac{32}{3}.$ Why this antiderivative: $\frac{d}{dx}(4x)=4$ and $\frac{d}{dx}(x^3/3)=x^2$. ✔

Intersections: $x^3=x\Rightarrow x^3-x=0\Rightarrow x(x-1)(x+1)=0$, so $x=-1,0,1$. There is a crossing inside the interval at $x=0$ — a warning sign. Order flips! Test $x=0.5$: $x=0.5$, $x^3=0.125$, so line is top on $(0,1)$. Test $x=-0.5$: $x=-0.5$, $x^3=-0.125$, so $x^3>x$ here, cube is top on $(-1,0)$. Split at the crossing and take (top$-$bottom) on each piece: $A={\int }_{-1}^0(x^3-x)dx+{\int }_0^1(x-x^3)dx.$ First: $[\frac{x^4}{4}-\frac{x^2}{2}{]}_{-1}^0=0-(\frac{1}{4}-\frac{1}{2})=\frac{1}{4}$. Second: $[\frac{x^2}{2}-\frac{x^4}{4}{]}_0^1=(\frac{1}{2}-\frac{1}{4})-0=\frac{1}{4}$. $A=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$

See s02. The parabola $x=y^2$ has a nose at $x=0$ and passes through $x=1$ (at $y=\pm 1$) and $x=4$ (at $y=\pm 2$). The line meets it at $(1,-1)$ and $(4,2)$. Scanning left to right, the top and bottom of a vertical strip change at $x=1$:

• For $0\le x\le 1$: strip runs from the lower parabola branch $y=-\sqrt{x}$ up to the upper branch $y=+\sqrt{x}$. Height $=\sqrt{x}-(-\sqrt{x})=2\sqrt{x}$.
• For $1\le x\le 4$: strip runs from the line $y=x-2$ (bottom) up to $y=+\sqrt{x}$ (top). Height $=\sqrt{x}-(x-2)$. $A={\int }_0^{1}2\sqrt{x}dx+{\int }_1^4(\sqrt{x}-x+2)dx.$ First: $[\frac{4}{3}{x}^{3/2}{]}_0^1=\frac{4}{3}$. Second: $[\frac{2}{3}{x}^{3/2}-\frac{x^2}{2}+2x{]}_1^4$. At $4$: $\frac{2}{3}\cdot 8-8+8=\frac{16}{3}$. At $1$: $\frac{2}{3}-\frac{1}{2}+2=\frac{13}{6}$. Difference $=\frac{16}{3}-\frac{13}{6}=\frac{32}{6}-\frac{13}{6}=\frac{19}{6}$. $A=\frac{4}{3}+\frac{19}{6}=\frac{8}{6}+\frac{19}{6}=\frac{27}{6}=\frac{9}{2}.✓$ Lesson: same answer, but two integrals and a branch-split versus the single $dy$ integral. Horizontal wins here.

Crossing inside? $\sin x=\cos x\Rightarrow \tan x=1\Rightarrow x=\frac{\pi }{4}$, which is inside $[0,\frac{\pi }{2}]$. Split there. Order on each piece: at $x=0$, $\cos 0=1>0=\sin 0$, so $\cos$ is top on $[0,\frac{\pi }{4}]$. At $x=\frac{\pi }{2}$, $\sin =1>0=\cos$, so $\sin$ is top on $[\frac{\pi }{4},\frac{\pi }{2}]$. $A={\int }_0^{\pi /4}(\cos x-\sin x)dx+{\int }_{\pi /4}^{\pi /2}(\sin x-\cos x)dx.$ First: $[\sin x+\cos x{]}_0^{\pi /4}=(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2})-(0+1)=\sqrt{2}-1$. Second: $[-\cos x-\sin x{]}_{\pi /4}^{\pi /2}=(0-1)-(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2})=-1+\sqrt{2}$. $A=(\sqrt{2}-1)+(\sqrt{2}-1)=2\sqrt{2}-2\approx \mathrm{0.828.}$

See s03. Get the three edge lines.

• $(0,0)\to (2,1)$: slope $\frac{1}{2}$, so $y=\frac{1}{2}x$.
• $(0,0)\to (1,3)$: slope $3$, so $y=3x$.
• $(2,1)\to (1,3)$: slope $\frac{3-1}{1-2}=-2$, so $y-1=-2(x-2)\Rightarrow y=-2x+5$. Scanning left to right the top edge changes at $x=1$ (the vertex $(1,3)$):
• $0\le x\le 1$: top $=y=3x$, bottom $=y=\frac{1}{2}x$. Height $=3x-\frac{1}{2}x=\frac{5}{2}x$.
• $1\le x\le 2$: top $=y=-2x+5$, bottom $=y=\frac{1}{2}x$. Height $=-2x+5-\frac{1}{2}x=-\frac{5}{2}x+5$. $A={\int }_0^1\frac{5}{2}xdx+{\int }_1^2(-\frac{5}{2}x+5)dx.$ First: $[\frac{5}{4}{x}^2{]}_0^1=\frac{5}{4}$. Second: $[-\frac{5}{4}{x}^2+5x{]}_1^2=(-5+10)-(-\frac{5}{4}+5)=5-\frac{15}{4}=\frac{5}{4}$. $A=\frac{5}{4}+\frac{5}{4}=\frac{5}{2}.$ Shoelace check: $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=\frac{1}{2}|0(1-3)+2(3-0)+1(0-1)|=\frac{1}{2}|6-1|=\frac{5}{2}.$ ✔

Why horizontal: the parabola is given as $x=f(y)$ already, so slice in $y$ — no inversion needed. Intersections: $6y-y^2=y\Rightarrow 5y-y^2=0\Rightarrow y(5-y)=0$, so $y=0,5$. Right/left: at $y=1$, parabola $=6-1=5$; line $=1$. So parabola is right, line is left. $A={\int }_0^5((6y-y^2)-y)dy={\int }_0^5(5y-y^2)dy=[\frac{5y^2}{2}-\frac{y^3}{3}{]}_0^5.$ At $y=5$: $\frac{5\cdot 25}{2}-\frac{125}{3}=\frac{125}{2}-\frac{125}{3}=\frac{375-250}{6}=\frac{125}{6}.$ $A=\frac{125}{6}\approx \mathrm{20.83.}$

Intersections: $x-x^2=mx\Rightarrow x-x^2-mx=0\Rightarrow x((1-m)-x)=0$, so $x=0$ and $x=1-m$. Top/bottom: for small $x>0$ the parabola $x-x^2\approx x$ exceeds $mx$ (since $m<1$), so parabola is top. $A(m)={\int }_0^{1-m}((x-x^2)-mx)dx={\int }_0^{1-m}((1-m)x-x^2)dx.$ Let $k=1-m$: ${\int }_0^k(kx-x^2)dx=[\frac{k{x}^2}{2}-\frac{x^3}{3}{]}_0^k=\frac{k^3}{2}-\frac{k^3}{3}=\frac{k^3}{6}.$ Set equal to $\frac{1}{6}$: $\frac{k^3}{6}=\frac{1}{6}\Rightarrow k^3=1\Rightarrow k=1\Rightarrow 1-m=1\Rightarrow m=0.$ That violates $m>0$, so no admissible $m$ gives area $\frac{1}{6}$; the maximum area (at $m\to 0^{+}$) is exactly $\frac{1}{6}$ and it is only approached, never reached for $m>0$. Answer: $A(m)=\frac{(1-m{)}^3}{6}$; area $=\frac{1}{6}$ requires $m=0$, so no valid $m>0$ exists — the supremum $\frac{1}{6}$ is not attained.

Vertical ($dx$). On $[0,1]$, at $x=\frac{1}{4}$: $\sqrt{x}=\frac{1}{2}>\frac{1}{4}=x$, so $\sqrt{x}$ is top. $A={\int }_0^1(\sqrt{x}-x)dx=[\frac{2}{3}{x}^{3/2}-\frac{x^2}{2}{]}_0^1=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}.$ Horizontal ($dy$). Invert (uses Inverse Functions): $y=\sqrt{x}\Rightarrow x=y^2$; $y=x\Rightarrow x=y$. For $y\in [0,1]$, at $y=\frac{1}{2}$: $x=y=\frac{1}{2}$ vs $x=y^2=\frac{1}{4}$, so the line $x=y$ is right, $x=y^2$ is left. $A={\int }_0^1(y-y^2)dy=[\frac{y^2}{2}-\frac{y^3}{3}{]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}.$ Why they must match: both integrals are Riemann sums (Definite Integral as Riemann Sum) of the same set of unit-area pieces tiling the same region; the direction of slicing only reorders the sum, and finite reordering of a convergent sum of areas cannot change the total. Hence $A=\frac{1}{6}$ both ways. $■$

Intersections: $y^2=y^2-2y\Rightarrow 0=-2y\Rightarrow y=0$. They meet at only one point, so they do not enclose a bounded region on their own — we need explicit $y$-bounds or another boundary. Re-read: the phrasing "between their intersections" is a trap since there is a single tangential meeting at $y=0$. Careful check of the gap: for $y\ne 0$, $R(y)-L(y)=y^2-(y^2-2y)=2y$. This is positive for $y>0$ and negative for $y<0$, meaning $x=y^2$ is right only for $y>0$; for $y<0$ the roles flip. With just these two curves and one crossing, no closed area exists — the answer is "$0$ enclosed area / region not bounded." Lesson: always confirm the curves actually enclose a region (need $\ge 2$ crossings for a lens) before integrating.

Point of technique ::: Ask "what changes as I sweep?" — if the top/bottom changes but right/left doesn't, slice horizontally. Before integrating ::: Confirm the curves cross at two points (real bounded region), then test an interior point for order. If curves cross inside ::: Split at every crossing; integrate (top$-$bottom) on each piece — never straight through. Sanity check ::: Match thickness to limits ($dx\to x$-limits, $dy\to y$-limits) and, when feasible, compute both ways.