Topic 8.5 Finding the Area Between Curves Expressed as Functions of y: integrate right minus left with respect to y to find the enclosed area.

What this topic is asking

The College Board (Topic 8.5) finds areas by integrating with respect to $y$ instead of $x$. When curves are more naturally written as $x = f(y)$, integrating right minus left with respect to $y$ often avoids splitting the region into pieces.

The method

A worked area in y

When y-integration avoids splitting

The practical reason to integrate in $y$ is that some regions are bounded above and below by the same curve when viewed in $x$, forcing a split, but are cleanly bounded left and right when viewed in $y$. A region enclosed by a sideways parabola $x = y^2$ and a vertical or slanted line is the classic case: integrating in $x$ would need two integrals (the parabola gives two $y$-branches), while a single integral in $y$ handles it. Recognizing when a region is "simpler sideways" saves work and reduces errors. The trade-off is that you must rewrite the bounding curves as $x$ in terms of $y$.

Choosing the orientation deliberately

The exam sometimes specifies "integrate with respect to $y$", and sometimes leaves the choice to you. When free to choose, pick the orientation that needs one integral rather than several: count how many times each vertical line (for $x$-integration) or horizontal line (for $y$-integration) crosses the boundary, and choose the direction where the region has a single top-and-bottom or single left-and-right description. The most common error in $y$-integration is keeping the curves as $y = \ldots$ and integrating in $y$ anyway; you must first solve for $x$ as a function of $y$, then integrate right minus left. Done correctly, $y$-integration is just the $x$-method with the roles of the axes swapped.

The horizontal-strip picture

The $y$-integration formula comes from slicing the region into thin horizontal strips. At a height $y$, a strip runs from the left curve $g(y)$ across to the right curve $f(y)$, so its width is $f(y) - g(y)$ and its area is that width times the thickness $dy$. Summing the strips from $y = c$ to $y = d$ gives $\int_c^d [f(y) - g(y)]\,dy$. Where vertical strips (in $x$) would cross the boundary an awkward number of times, horizontal strips may cross it just twice, giving a single clean integral. Visualizing the strips horizontal rather than vertical is the mental switch that makes $y$-integration natural.

A worked comparison of orientations

Consider the region bounded by $y = \sqrt{x}$, the $x$-axis, and $x = 4$. Integrating in $x$, vertical strips run from the axis up to $y = \sqrt{x}$, giving $\int_0^4 \sqrt{x}\,dx$, one clean integral. Integrating in $y$ instead, horizontal strips run from the curve $x = y^2$ across to $x = 4$, giving $\int_0^2 (4 - y^2)\,dy$, also one integral. Both yield the same area, $\frac{16}{3}$, so here either orientation works. The orientation only becomes decisive when one direction would force a split; recognizing in advance which way avoids the split is what the deliberate-choice skill is really about, and trying a quick strip count in each direction settles it.