Choosing dx or dy

A region can be cut into vertical strips, which gives an integral in

x

of top minus bottom, or into horizontal strips, which gives an integral in

y

of right minus left. The two integrals give the same area, and often one of them is much less work.

The same region in x and in y

Example 1

Find the area of the region in the first quadrant bounded by the curve

y = \ln x

, the line

y = 1

, and the two coordinate axes.

The curve crosses the

x

-axis at

(1, 0)

and meets the line

y = 1

(e, 1)

. The corners of the region are

(0, 0)

(1, 0)

(e, 1)

and

(0, 1)

Vertical strips run up to y = 1 from the x-axis left of the dashed line x = 1, and from the curve right of it (one is drawn). A horizontal strip runs from the y-axis to the curve.

With respect to $x$ . The top is the line

y = 1

throughout, but the bottom is the

x

-axis for

0 \le x \le 1

and the curve for

1 \le x \le e

, so the integral splits:

A = \int_{0}^{1} 1\,dx + \int_{1}^{e} \left(1 - \ln x\right)dx.

None of the basic antiderivative rules covers

\ln x

. Integration by parts, a BC topic, gives

x\ln x - x

(its derivative is

\ln x + 1 - 1

, which checks it), so an antiderivative of

1 - \ln x

x - (x\ln x - x) = 2x - x\ln x

. Then

A = 1 + \left(2x - x\ln x\right)\Big|_{1}^{e} = 1 + (2e - e) - (2 - 0) = e - 1.

With respect to $y$ . Solving

y = \ln x

for

x

gives

x = e^y

. A horizontal strip at height

y

runs from the

y

-axis on the left to the curve on the right, for

y

from

0

1

A = \int_{0}^{1} e^{y}\,dy = e^{y}\Big|_{0}^{1} = e - 1.

Both ways give

e - 1 \approx 1.718

. In

x

the region takes two integrals and an antiderivative of

\ln x

. In

y

it takes one exponential integral.

Avoiding a split

A split comes from a boundary that changes formula partway across. Before choosing a variable, sketch the region and find its corners. The sketch shows which boundaries each kind of strip touches.

Example 2

Find the area of the region bounded by the line

y = x

, the curve

y = \sqrt[3]{x}

, and the line

y = 2

. Its lowest corner is the crossing at

(1, 1)

The line

y = 2

meets

y = x

(2, 2)

and meets

y = \sqrt[3]{x}

(8, 2)

. The corners are

(1, 1)

(2, 2)

and

(8, 2)

Left of the dashed line x = 2 the top of the region is y = x; right of it, y = 2. A horizontal strip runs from the line y = x to the curve.

With respect to $x$ . The bottom is the curve throughout, but the top changes at

x = 2

, so the integral splits:

\begin{aligned} A &= \int_{1}^{2} \left(x - \sqrt[3]{x}\right)dx + \int_{2}^{8} \left(2 - \sqrt[3]{x}\right)dx \\[4pt] &= \left(\frac{x^2}{2} - \frac{3}{4}\,x^{4/3}\right)\Big|_{1}^{2} + \left(2x - \frac{3}{4}\,x^{4/3}\right)\Big|_{2}^{8} \\[4pt] &= \left(\frac{9}{4} - \frac{3}{4}\cdot 2^{4/3}\right) + \frac{3}{4}\cdot 2^{4/3} = \frac{9}{4}. \end{aligned}

With respect to $y$ . The line is

x = y

and the curve is

x = y^3

. Every horizontal strip runs from the line on the left to the curve on the right, for

y

from

1

2

, so one integral covers the region:

\begin{aligned} A &= \int_{1}^{2} \left(y^3 - y\right)dy = \left(\frac{y^4}{4} - \frac{y^2}{2}\right)\Big|_{1}^{2} \\[4pt] &= (4 - 2) - \left(\frac{1}{4} - \frac{1}{2}\right) = \frac{9}{4}. \end{aligned}

When x is the only choice

An integral in

y

needs every boundary written as

x

in terms of

y

. When a curve can't be solved for

x

, there is no formula to integrate in

y

, and vertical strips are the way to find the area.

Example 3

Find the area of the region bounded by

y = x + \sin x

, the

x

-axis, and the line

x = \pi

The region under y = x + sin x from x = 0 to x = π, with one vertical strip.

The curve passes through the origin, and

x + \sin x \ge 0

for

0 \le x \le \pi

. Vertical strips run from the

x

-axis up to the curve:

\begin{aligned} A &= \int_{0}^{\pi} (x + \sin x)\,dx \\[4pt] &= \left(\frac{x^2}{2} - \cos x\right)\Big|_{0}^{\pi} \\[4pt] &= \left(\frac{\pi^2}{2} + 1\right) - (0 - 1) = \frac{\pi^2}{2} + 2 \approx 6.935. \end{aligned}

With horizontal strips, the left end of each strip is the point of the curve at height

y

, and that needs the solution of

x + \sin x = y

for

x

. The curve is continuous and increasing, so each height has exactly one solution, but no algebra gives a formula for it. An integral in

y

would have nothing to find an antiderivative of.

Rewriting an integral in y as one in x

To switch variables, draw the region, then describe it again with the other kind of strip. The integrand alone doesn't fix the region (it is only the length of a strip), so start from the boundaries. The limits change too: an integral in

x

runs between

x

-values, not

y

-values.

Example 4

The region

R

lies between the curve

x = \sqrt{y}

and the line

x = 2

, for

1 \le y \le 4

. Its area is

\int_{1}^{4} \left(2 - \sqrt{y}\right)dy.

(a) Find the corners of

R

. (b) Write the area of

R

as an integral with respect to

x

. (c) Evaluate both integrals.

(a) The curve meets the line

x = 2

y = 4

. At the bottom,

y = 1

, the strip runs from

x = 1

x = 2

. The corners are

(1, 1)

(2, 1)

and

(2, 4)

, and the bottom edge is the line

y = 1

R, with a horizontal strip from x = √y to x = 2 and a vertical strip from y = 1 to y = x².

(b) For

x \ge 0

, the curve

x = \sqrt{y}

y = x^2

. A vertical strip runs from the line

y = 1

up to the curve, and

x

runs from

1

2

A = \int_{1}^{2} \left(x^2 - 1\right)dx.

(c) Both integrals give the same value:

\begin{aligned} \int_{1}^{4} \left(2 - \sqrt{y}\right)dy &= \left(2y - \frac{2}{3}\,y^{3/2}\right)\Big|_{1}^{4} = \frac{8}{3} - \frac{4}{3} = \frac{4}{3}, \\[4pt] \int_{1}^{2} \left(x^2 - 1\right)dx &= \left(\frac{x^3}{3} - x\right)\Big|_{1}^{2} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}. \end{aligned}