Finding the Area Between Curves Expressed as Functions of y

AP Calculus · Applications of Integration

Area Between Curves as Functions of y

Rotate the whole method a quarter turn: horizontal slices, right minus left, dy.

key idea$A=\int_c^d\Big[\text{right}(y)-\text{left}(y)\Big]dy$

Picture the region

Everything from vertical slicing carries over, rotated a quarter turn. The slice is now horizontal, its length is $\text{right}-\text{left}$, and its thickness is $dy$. The limits are $y$-values, and every boundary must be rewritten as $x$ in terms of $y$.

The region bounded by $x=y^{2}$ and $x=y+2$, sliced horizontally. The slice starts on the parabola and ends on the line, so its length is $(y+2)-y^{2}$, and the slices run from $y=-1$ up to $y=2$.

The same region, sliced vertically instead. Now the bottom boundary changes at $x=1$ — the rose line — from the lower half of the parabola to the line. That single switch forces the area into two integrals, which is exactly what horizontal slicing avoids.

Setting up the integral

Slicing horizontally

Solve every boundary for $x$ as a function of $y$. A curve given as $y=\sqrt{x}$ becomes $x=y^{2}$; a line $y=2x+1$ becomes $x=\dfrac{y-1}{2}$.
Find the limits by setting the two $x$-expressions equal and solving for $y$. The limits are heights, not horizontal positions.
Test one interior $y$ to see which curve is further right. Substitute it into both.
Write $\text{right}-\text{left}$ and integrate with respect to $dy$ over $[c,d]$.
Check the answer is positive. As with vertical slicing, a negative result means the two boundaries were subtracted the wrong way round.

Choose horizontal slices when…	Because…
the region is bounded left and right by curves	each slice has one clear start and one clear end
the bottom boundary changes partway across	vertical slicing would need two integrals
the curves are naturally $x=g(y)$	no rearranging is needed at all
a boundary is a vertical line $x=k$	it becomes a constant limit for every slice
solving for $y$ produces a $\pm$	the two branches would have to be handled separately

The decision is purely practical: pick the direction that needs fewer integrals. Both give the same number, so the only thing at stake is how much work it takes to get there.

Worked examples

EX 1

One integral instead of two

$x=y^{2}$ and $x=y+2$

Find the area bounded by $x=y^{2}$ and $x=y+2$.

Find the limits in $y$

$y^{2}=y+2$ gives $y^{2}-y-2=0$, so $(y-2)(y+1)=0$ and $y=-1,\,2$.

Decide which is on the right

At $y=0$: the line gives $x=2$, the parabola gives $x=0$. The line is on the right.

Write the integral

$A=\displaystyle\int_{-1}^{2}\left[(y+2)-y^{2}\right]dy$.

Antidifferentiate

$\left[\dfrac{y^{2}}{2}+2y-\dfrac{y^{3}}{3}\right]_{-1}^{2}$.

Evaluate

$\left(2+4-\tfrac83\right)-\left(\tfrac12-2+\tfrac13\right)=\dfrac{10}{3}+\dfrac76=\dfrac92$.

EX 2

Rewriting the boundaries

solve for $x$ first

Find the area between $y=\sqrt{x}$, the $y$-axis, and $y=3$.

Rewrite in terms of $y$

$y=\sqrt{x}$ becomes $x=y^{2}$; the $y$-axis is $x=0$.

Identify right and left

For $0\lt y\lt3$, the curve $x=y^{2}$ is to the right of $x=0$.

Set the limits

The region runs from $y=0$ to $y=3$.

Integrate

$A=\displaystyle\int_{0}^{3}\left(y^{2}-0\right)dy=\left[\dfrac{y^{3}}{3}\right]_{0}^{3}=9$.

Notice the simplification

Sliced vertically this would need $\int\left(3-\sqrt{x}\right)dx$ over $[0,9]$ — correct, but a messier antiderivative for the same answer.

EX 3

A vertical line as a boundary

constant limit

Find the area bounded by $x=y^{2}$ and $x=4$.

Find the limits

$y^{2}=4$ gives $y=\pm2$.

Identify right and left

At $y=0$: $x=4$ versus $x=0$. The vertical line is on the right.

Write the integral

$A=\displaystyle\int_{-2}^{2}\left(4-y^{2}\right)dy$.

Use the symmetry

The integrand is even, so $A=2\displaystyle\int_{0}^{2}\left(4-y^{2}\right)dy$.

Evaluate

$2\left[4y-\dfrac{y^{3}}{3}\right]_{0}^{2}=2\left(8-\dfrac83\right)=\dfrac{32}{3}$.

EX 4

Deciding which way to slice

count the integrals

The region is bounded by $y=x-2$, $y=-\sqrt{x}$ and $y=\sqrt{x}$. Which slicing direction is better?

Try vertical slices

For $0\le x\le1$ the bottom is $y=-\sqrt{x}$; for $1\le x\le4$ it is $y=x-2$. The bottom changes.

Count the integrals

Two, split at $x=1$, plus the work of finding that split point.

Try horizontal slices

Rewrite: $x=y^{2}$ on the left, $x=y+2$ on the right, for all $y$ from $-1$ to $2$.

Count again

One integral, no split.

Choose

Slice horizontally. The rule of thumb: whichever direction keeps a single pair of boundaries throughout is the right one.

Watch out

Every boundary must be $x$ in terms of $y$

Leaving one curve as $y=f(x)$ makes the subtraction meaningless. Convert all of them before writing the integral.

The limits are $y$-values

Solve the boundary equations for $y$, not $x$. Using $x$-limits with a $dy$ integral is the most common set-up error here.

Right minus left, and the differential is $dy$

Both halves must switch together. A $\text{right}-\text{left}$ integrand with $dx$ attached is not an area.

Pick the direction with fewer integrals

Both directions are correct. If one boundary changes partway along, slicing the other way usually removes the split entirely.

Course map. Topic 8.5 · LO CHA-5.A — calculate areas in the plane using the definite integral.