Choosing the slice direction

In every area problem you have a choice: integrate by sweeping a vertical bar from left to right (using $dx$ as the slice width), or by sweeping a horizontal bar from bottom to top (using $dy$). The answer is always the same, but the work involved can differ wildly. A region that needs a delicate splitting under one orientation may be a one-line integral under the other.

When vertical ($dx$) slices win

Use $dx$-slices when:

• The curves are easily written as $y = f(x)$, $y = g(x)$.
• The region's left and right boundaries are vertical lines ($x = a$ and $x = b$), or two curves whose $x$-intersections give clean limits.
• The top and bottom boundaries don't change as $x$ varies through the interval.

The typical setup: a region between $y = f(x)$ and $y = g(x)$ over $[a, b]$, with $f \ge g$. Area $= \int_a^b [f - g]\,dx$.

When horizontal ($dy$) slices win

Use $dy$-slices when:

• Curves are easier as $x = u(y)$, $x = v(y)$. Classic examples: $y^2 = 4ax$ (parabola opening right), $x = y^2$, and similar.
• The region's top and bottom are horizontal.
• A horizontal sweep keeps the same left/right boundaries throughout, but a vertical sweep would require splitting.

The setup: a region between $x = u(y)$ (right) and $x = v(y)$ (left) over $[c, d]$. Area $= \int_c^d [u - v]\,dy$.

When splitting is unavoidable

Some regions have boundaries that change at some interior point , e.g. a region whose top boundary is one curve up to a vertex and another curve afterwards. Then you split the integral at that point, regardless of slice direction.

If both vertical and horizontal sweeps need splitting, choose whichever requires fewer splits or simpler pieces.

Translating between slice directions

A region described by "between $y = f(x)$ and the $x$-axis over $[a, b]$" can also be described by horizontal slices. At height $y \in [0, M]$ where $M$ is the maximum of $f$ on $[a, b]$, the slice extends horizontally from the leftmost $x$ with $f(x) = y$ to the rightmost. If $f$ is invertible, this becomes $x = f^{-1}(y)$.

Practical translation: sketch, identify horizontal slices, write their endpoints in terms of $y$, integrate.

Worked examples

Example 1. Region between $y^2 = x$, $y = 0$, $y = 2$.

Vertical sweep: at each $x \in [0, 4]$, slice from $y = 0$ to $y = \sqrt x$ (upper branch). But there's also a piece for $x \in [0, 4]$ from $y = -\sqrt x$ to $y = 0$ , except $y = 0$ to $y = 2$ excludes the negative-$y$ branch. So just one piece: $\int_0^4 \sqrt x\,dx = \dfrac{16}{3}$.

Horizontal sweep: at each $y \in [0, 2]$, slice from $x = 0$ to $x = y^2$. Area $= \int_0^2 y^2\,dy = \dfrac{8}{3}$.

Wait , these don't match. Let me re-read the problem. The region bounded by $y^2 = x$, $y = 0$, $y = 2$: this is bounded on the right by the parabola, below by $y = 0$, above by $y = 2$. The region is to the left of the parabola, between $y = 0$ and $y = 2$. At height $y$, the slice runs from $x = 0$ (the $y$-axis is the implicit left boundary) to $x = y^2$. So horizontal slice: area $= \int_0^2 y^2\,dy = \dfrac{8}{3}$. Vertical interpretation: at each $x \in [0, 4]$, the strip goes from $y = \sqrt x$ to $y = 2$. Area $= \int_0^4 (2 - \sqrt x)\,dx = 8 - \dfrac{16}{3} = \dfrac{8}{3}$. Now they match.

This example shows that the horizontal sweep is faster here.

Example 2. Region enclosed by $y = x^2$ and $y = x$.

Vertical sweep (preferred): $\int_0^1 (x - x^2)\,dx = 1/6$.

Horizontal sweep: at height $y \in [0, 1]$, the region runs from $x = y$ (line) to $x = \sqrt y$ (parabola). Area $= \int_0^1 (\sqrt y - y)\,dy = \dfrac{2}{3} - \dfrac{1}{2} = 1/6$. Both work; the vertical is slightly simpler.

Example 3. Region enclosed by $x = y^2$, $x = 1$.

Horizontal sweep (preferred): at height $y \in [-1, 1]$, $x$ goes from $y^2$ to $1$. Area $= \int_{-1}^1 (1 - y^2)\,dy = 2 - 2/3 = 4/3$.

Vertical sweep: at each $x \in [0, 1]$, $y$ goes from $-\sqrt x$ to $\sqrt x$, so the slice has height $2\sqrt x$. Area $= \int_0^1 2\sqrt x\,dx = 4/3$. Both work; horizontal slightly cleaner here.

Example 4. Region bounded by $y = \ln x$, $y = 0$, $x = e$.

Vertical sweep: $\int_1^e \ln x\,dx = [x\ln x - x]_1^e = (e - e) - (-1) = 1$.

Horizontal sweep: at height $y \in [0, 1]$, $x$ runs from $e^y$ (the inverse function) to $e$. Area $= \int_0^1 (e - e^y)\,dy = e - (e - 1) = 1$. Both yield 1.

Example 5. Region enclosed by $y = x^3$ and $y = x$.

Two pieces by symmetry, each in one quadrant. Vertical sweep: $\int_0^1 (x - x^3)\,dx = 1/4$, doubled to $1/2$. Horizontal sweep: in the first quadrant, at height $y \in [0, 1]$, $x$ runs from $y$ (line) to $y^{1/3}$ (cube-root curve, since $y = x^3 \Leftrightarrow x = y^{1/3}$). Area $= \int_0^1 (y^{1/3} - y)\,dy = 3/4 - 1/2 = 1/4$. Same answer.

Example 6. Region bounded by $y = e^x$, $y = 2$, $x = 0$.

Vertical sweep: at $x \in [0, \ln 2]$, $y$ runs from $e^x$ to $2$. Area $= \int_0^{\ln 2} (2 - e^x)\,dx = 2\ln 2 - (2 - 1) = 2\ln 2 - 1$.

Horizontal sweep: at height $y \in [1, 2]$, $x$ runs from $0$ to $\ln y$. Area $= \int_1^2 \ln y\,dy = [y \ln y - y]_1^2 = (2\ln 2 - 2) - (0 - 1) = 2 \ln 2 - 1$. Same.

Try it yourself

For each region below, set up area integrals both ways (using $dx$ and using $dy$), compute one, verify with the other.

1. Region bounded by $y = \sqrt x$, $y = 0$, $x = 4$.
2. Region bounded by $y = x^2$, $x = 1$, $y = 0$.
3. Region bounded by $y^2 = 4x$, $x = 1$ (in the upper half).
4. Region bounded by $y = \cos x$, $y = 0$, $x = 0$, $x = \pi/2$.
5. Region bounded by $x = y^2 - 1$ and $x = 0$.
6. Region bounded by $y = x$, $y = 2x$, $x = 1$.
7. Region bounded by $x^2 + y^2 = 4$ in the first quadrant.
8. Region bounded by $y = e^{-x}$, $y = 0$, $x = 0$, $x = 1$.
9. Region bounded by $y = x^2 - 4x + 3$ and the $x$-axis.
10. Region bounded by $y = \sin x$ on $[0, \pi]$ and the $x$-axis.
11. Region in the first quadrant bounded by $y = x^2$, $y = 4$.
12. Region bounded by $y = \tan^{-1}x$, $y = 0$, $x = 1$.
13. Region bounded by $y^2 = 2x$ and $y = x - 4$.
14. Region between $y = 2x$ and $y = x^2$ in the first quadrant.

Pitfalls and tricks

• Sketch first. Without a sketch you cannot decide which sweep is cleaner.
• Inverse functions and slice direction. Switching slice direction usually means writing the inverse , make sure that inverse exists on the relevant part of the curve.
• One-piece integrals beat multi-piece ones. If $dx$ requires splitting and $dy$ doesn't (or vice versa), choose the one without splits.
• For parabolas opening sideways, horizontal slices are almost always cleaner.
• For parabolas opening up/down, vertical slices usually win.