Area between curves and changing orientation

Understand the idea

Find all intersections that bound the region and identify which curve is above or to the right on each subinterval. Vertical slices integrate top minus bottom with respect to $\displaystyle x$ ; horizontal slices integrate right minus left with respect to $\displaystyle y$ . Split where the ordering changes. Geometric area is nonnegative, so an unsplit signed integral may not answer the question.

A=\int_a^b(f(x)-g(x))\,dx\quad(f\ge g),\qquad A=\int_c^d(x_R(y)-x_L(y))\,dy

Worked example

Find the area between $\displaystyle y=x$ and $\displaystyle y=x^2$ from their intersections.

Solve

\displaystyle x=x^2

to get

\displaystyle 0,1

. On

\displaystyle (0,1)

the line is above the parabola, so area is

\displaystyle \int_0^1(x-x^2)\,dx=1/2-1/3=1/6

.

Try it yourself

Find the area between $\displaystyle x=y^2$ and $\displaystyle x=2y$ in their bounded region.

Show solution and reasoning

The intersections have $\displaystyle y=0,2$ . Right minus left is $\displaystyle 2y-y^2$ , giving $\displaystyle \int_0^2(2y-y^2)\,dy=4/3$ .