Mistake Master

Finding the Area Between Curves Expressed as Functions of y AB & BC

A horizontal strip at height $y$ has length $x_{\text{right}} - x_{\text{left}}$ and thickness $dy$, so the area is $\int_c^d (x_{\text{right}} - x_{\text{left}})\,dy$ with $y$-values as the limits. This is 8.4 rotated: top becomes right, $dx$ becomes $dy$, and every check carries over unchanged. The region between $x = y + 2$ and $x = y^{2}$ is the reflection of 8.4's region across $y = x$ and has the same area $\frac{9}{2}$, found from $y^{2} = y + 2$ with $y = -1$ and $y = 2$ and a test at $y = 0$ that puts the line on the right.

Every boundary must be rewritten as $x = g(y)$ before integrating, the branch restriction travels with it, and the limits have to be recomputed as the $y$-coordinates of the intersection points rather than reused from a $dx$ setup. A surviving $x$ anywhere in a $dy$ integral means a boundary was never converted. Which variable to sweep in is decided by how often a boundary hands off to a different curve: the region between $x = y^{2}$ and $x = y + 2$ needs one integral in $y$ and two in $x$, since the upper boundary changes from the parabola's upper branch to the line at $x = 1$.

A HORIZONTAL STRIP RUNS FROM THE PARABOLA TO THE LINE. x = y + 2 ON THE RIGHT x = y² ON THE LEFT THEY MEET AT y = −1 AND AT y = 2 STRIP LENGTH IS (y + 2) − y² AREA = 9/2 THE LIMITS ARE THE y-COORDINATES OF THE TWO MARKED POINTS, NOT THEIR x-COORDINATES 1 AND 4.
Drawn to scale at $62$ px per unit horizontally and $48$ px per unit vertically, with the parabola computed rather than sketched. The marked strip sits at $y = 0.8$, where it runs from $x = 0.64$ to $x = 2.8$. This is 8.4's figure reflected across the line $y = x$.
THE SAME REGION SWEPT THE OTHER WAY NEEDS TWO INTEGRALS. THE DASHED LINE IS x = 1, WHERE THE TOP BOUNDARY CHANGES LEFT STRIP: PARABOLA TO PARABOLA RIGHT STRIP: LINE TO PARABOLA SWEEPING SIDEWAYS, EVERY STRIP GOES PARABOLA TO LINE: ONE INTEGRAL. THE COUNT IS THE ONLY REASON TO PREFER ONE VARIABLE.
The two marked vertical strips sit at $x = 0.5$ and $x = 2.5$. The left one runs between the two branches of the parabola; the right one runs from the line up to the parabola. Both integrals are needed, and both together give the same $\frac{9}{2}$ that one $dy$ integral gives.

The work

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Lesson
Finding the Area Between Curves Expressed as Functions of y

Rotates 8.4 rather than replacing it: a horizontal strip of length right minus left, thickness dy, and y-values as limits. Converts every boundary to x as a function of y with its branch restriction, recomputes the limits from the intersection points, and settles which variable costs fewer integrals.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items on area between curves in y: converting boundaries to functions of y, using y-coordinates rather than x-coordinates as limits, ordering right minus left by testing a point, and choosing the sweep direction that avoids a split.

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