Finding the Area Between Curves Expressed as Functions of y
Introduced| English |
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| horizontal slice/ˌhɒrɪˈzɒntl slaɪs/ |
Slice sideways instead of up-and-down
- Some regions are awkward with vertical strips — a sideways-opening parabola, say.
- The fix: slice into horizontal strips and integrate with respect to $y$ instead of $x$.
- Everything from the last lesson turns 90°: curves become functions of $y$, and you subtract right minus left.
- Same idea, rotated: add up horizontal gap-widths.
Right minus left
- Write the boundaries as functions of $y$: $x=f(y)$ (rightmost) and $x=g(y)$ (leftmost).
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$$A=\int_c^d \big(f(y)-g(y)\big)\,dy$$
- Each horizontal slice 水平切片 has width $f(y)-g(y)$ and height $dy$.
- Subtract rightmost minus leftmost so the width stays positive.
A sideways-opening parabola
y = ax² + bx + c
When curves open sideways, horizontal slices (right minus left, integrated in $y$) are the clean choice.
With horizontal slices, the area is $\int_c^d(\ ?\ )\,dy$ where...
Horizontal strips: right curve minus left curve.
Bounds are now $y$-values
- The limits $c$ and $d$ are the lowest and highest $y$ of the region — often the curves' intersection $y$-values.
- Solve $f(y)=g(y)$ (in terms of $y$) for those bounds.
- Between them, test which curve is farther right.
- Then integrate the right-minus-left width from $c$ to $d$.
For a $dy$ integral, the limits of integration are...
Integrating in $y$ means $y$-limits.
Where do $x=y^2$ and $x=y+2$ intersect (in $y$)?
$y^2-y-2=0\Rightarrow y=-1,2$.
When to choose $dy$ over $dx$
- Prefer horizontal slices when the curves are naturally $x=$(function of $y$), or when vertical slices would need splitting.
- A region bounded on the left and right by different curves (but top/bottom by single points) is a $dy$ situation.
- If the "upper" curve would change midway with vertical slices, horizontal slices may avoid the split.
- Pick whichever orientation makes each strip span one clean curve to one clean curve.
The area bounded by $x=y^2$ and $x=y+2$ is $\int_{-1}^2 (y+2-y^2)\,dy$. Compute it (a decimal).
It evaluates to $\tfrac92=4.5$.
For a horizontal-slice integral, you should write each boundary as $x$ in terms of $y$.
The strips run horizontally, so express $x$ as a function of $y$.
Horizontal slices are preferable when...
Sideways curves suit $dy$ integration.
With horizontal slices you integrate right minus left in terms of $y$, and the limits are $y$-values (not $x$-values). Mixing them up — using $x$-limits with a $dy$ integral, or subtracting top-minus-bottom sideways — gives a wrong answer. Rewrite each boundary curve as $x$ in terms of $y$ first.
Find the area bounded by $x=y^2$ and $x=y+2$.
- Intersections (in $y$): $y^2=y+2\Rightarrow y^2-y-2=0\Rightarrow y=-1,2$.
- On $(-1,2)$ the line $x=y+2$ is farther right (test $y=0$: $2>0$).
- $A=\displaystyle\int_{-1}^{2}\big((y+2)-y^2\big)\,dy=\tfrac{9}{2}$.
For curves given as functions of $y$, use horizontal slices: $A=\int_c^d\big(f(y)-g(y)\big)\,dy$, integrating rightmost minus leftmost with $y$-limits from the intersections. Choose $dy$ when it lets each strip run cleanly from one curve to another without splitting.