Finding the Area Between Curves

How do you find the area of a region bounded by two curves? I'll consider two cases.

Suppose the region is bounded above and below by the two curves("top" and "bottom"), and on the sides by $x = a$ and $x = b$ .

$$\hbox{\epsfysize=2in \epsffile{areacu1.eps}}$$

Think of dividing the region up into vertical rectangles. The height of the typical rectangle is $(\hbox{top}) - (\hbox{bottom})$ , while the thickness is $dx$ . The area of a typical rectangle is

$$\left((\hbox{top}) - (\hbox{bottom})\right)\,dx.$$

To find the total area, integrate to add up the areas of the little rectangles:

$$A = \int_a^b \left((\hbox{top}) - (\hbox{bottom})\right)\,dx.$$

The $dx$ in the integral is a reminder that I want "top" and "bottom" expressed in terms of x.

Suppose the region is bounded on the sides by two curves ("left" and "right"), and on the top and bottom by $y = c$ and $y = d$ .

$$\hbox{\epsfysize=2in \epsffile{areacu2.eps}}$$

Think of dividing the region up into horizontal rectangles. The height of the typical rectangle is $(\hbox{right}) - (\hbox{left})$ , while the thickness is $dy$ . The area of a typical rectangle is

$$\left((\hbox{right}) - (\hbox{left})\right)\,dy.$$

To find the total area, integrate to add up the areas of the little rectangles:

$$A = \int_c^d \left((\hbox{right}) - (\hbox{left})\right)\,dy.$$

The $dy$ in the integral is a reminder that I want "right" and "left" expressed in terms of y.

Example. Find the area of the region bounded by $y = 4 - x^2$ and the x-axis.

$$\hbox{\epsfysize=1.75in \epsffile{areacu3.eps}}$$

$$A = \int_{-2}^2 (4 - x^2)\,dx = \dfrac{32}{3}.\quad\halmos$$

Example. Find the area of the region bounded by $y = x^2$ and $y = x + 12$ .

$$\hbox{\epsfysize=1.75in \epsffile{areacu4.eps}}$$

$$A = \int_{-3}^4 (x + 12 - x^2)\,dx = \dfrac{343}{6}.\quad\halmos$$

Example. Find the area of the region bounded by $x = y^2 - 2y - 30$ and $x = 2y - y^2$ .

$$\hbox{\epsfysize=1.75in \epsffile{areacu5.eps}}$$

$$A = \int_{-3}^5 \left((2y - y^2) - (y^2 - 2y - 30)\right)\,dy = \dfrac{512}{3}.\quad\halmos$$

Example. Find the area of the region bounded by $y = x^3 - x$ and $y = -0.5 + x - x^3$ .

The curves intersect at the (approximate) values -1.10710, 0.83757, and 0.26959.

$$\hbox{\epsfysize=1.75in \epsffile{areacu6.eps}}$$

$$A = \int_{-1.10710}^{0.26959} \left((x^3 - x) - (-0.5 + x - x^3)\right)\,dx + \int_{0.26959}^{0.83757} \left((-0.5 + x - x^3) - (x^3 - x)\right)\,dx \approx 0.65077.\quad\halmos$$

Example. Find the area of the region between the curves $y = x^2$ and $y = 2 - x$ from $x = 0$ to $x = 2$ .

$$\hbox{\epsfysize=1.75in \epsffile{areacu7.eps}}$$

$$A = \int_0^1 (2 - x - x^2)\,dx + \int_1^2 (x^2 - (2 - x))\,dx = 3.\quad\halmos$$

Example. Find the area of the region between the curves $y = x^2 + 3$ and $y = \sqrt{1 - x^2}$ from $x = -1$ to $x = 1$ .

$$\hbox{\epsfysize=1.75in \epsffile{areacu8.eps}}$$

The area of the region under $y = x^2 + 3$ is

$$\int_{-1}^1 (x^2 + 3)\,dx = \left[\dfrac{1}{3}x^3 + 3x\right]_{-1}^1 = \dfrac{20}{3}.$$

The area of the semicircle is $\dfrac{\pi}{2}$ .

Hence, the area of the region between the curves is

$$A = \dfrac{20}{3} - \dfrac{\pi}{2} \approx 5.09587.\quad\halmos$$

Example. Find the area of the region bounded by $y = x^3 - 6x^2 - 16x$ and $y = 8x + 2x^2 - x^3$ .

The region consists of two pieces. For the left-hand piece, the top curve is $y = x^3 - 6x^2 - 16x$ and the bottom curve is $y = 8x + 2x^2 - x^3$ . For the right-hand piece, the top curve is $y = 8x + 2x^2 - x^3$ and the bottom curve is $y = x^3 - 6x^2 - 16x$ .

$$\hbox{\epsfysize=1.75in \epsffile{areacu10.eps}}$$

I need to find where the curves intersect. Solve the equations simultaneously:

$$x^3 - 6x^2 - 16x = 8x + 2x^2 - x^3, \quad 2x^3 - 8x^2 - 24x = 0, \quad x^3 - 4x^2 - 12x = 0,$$

$$x(x^2 - 4x - 12) = 0, \quad x(x - 6)(x + 2) = 0.$$

The intersections are at $x = -2$ , $x = 0$ , and $x = 6$ .

The area is

$$\int_{-2}^0 \left((x^3 - 6x^2 - 16x) - (8x + 2x^2 - x^3)\right)\,dx + \int_0^6 \left((8x + 2x^2 - x^3) - (x^3 - 6x^2 - 16x)\right)\,dx =$$

$$\int_{-2}^0 (2x^3 - 8x^2 - 24x)\,dx + \int_0^6 (-2x^3 + 8x^2 + 24x)\,dx =$$

$$\left[\dfrac{1}{2}x^4 - \dfrac{8}{3}x^3 - 12x^2\right]_{-2}^0 + \left[-\dfrac{1}{2}x^4 + \dfrac{8}{3}x^3 + 12x^2\right]_0^6 = \dfrac{1136}{3} \approx 378.66667.\quad\halmos$$

Example. Find the area of the region bounded above by the curves $y = x^2$ and $y = 6 - x$ and below by the x-axis:

(a) Using vertical rectangles.

(b) Using horizontal rectangles.

$$\hbox{\epsfysize=1.75in \epsffile{areacu9.eps}}$$

The curves intersect at $x = 2$ , $y = 4$ .

Using vertical rectangles, I need two integrals:

$$A = \int_0^2 x^2\,dx + \int_2^6 (6 - x)\,dx = \dfrac{32}{3}.$$

Using horizontal rectangles, I only need one integral:

$$A = \int_0^4 (6 - y - \sqrt{y})\,dy = \dfrac{32}{3}.\quad\halmos$$

Send comments about this page to: bikenaga@marauder.millersville.edu.

Last updated: December 2, 2005

Bruce Ikenaga's Home Page

Math Department Home Page

Millersville University Home Page

Copyright 2005 by Bruce Ikenaga