# Calculus

## Double Integrals # Double Integrals over General Regions

If a region R of type I is bounded by x = a, x = b, y = p(x) and y = q(x) with a < b and p(x) < q(x) for all x [a, b], then using the Fubini's theorem the double integral can be written as

$\iint\limits_R {f\left( {x,y} \right)dxdy} = \int\limits_{x = a}^{x = b} {\int\limits_{y = p\left( x \right)}^{y = q\left( x \right)} {f\left( {x,y} \right)dydx} } .$

The similar formula exists for regions of type $$II.$$ If a region $$R$$ of type $$II$$ is bounded by the graphs of the functions $$x = u\left( y \right),$$ $$x = v\left( y \right),$$ $$y = c,$$ $$y = d$$ provided that $$c \lt d$$ and $$u\left( y \right) \lt v\left( y \right)$$ for all $$y \in \left[ {c,d} \right],$$ then the double integral over the region $$R$$ is expressed through the iterated integral by the Fubini's theorem

$\iint\limits_R {f\left( {x,y} \right)dxdy} = \int\limits_{y = c}^{y = d} {\int\limits_{x = u\left( y \right)}^{x = v\left( y \right)} {f\left( {x,y} \right)dxdy} } .$

It is sometimes useful to break the region $$R$$ up into two or more smaller regions, and integrate over each separately.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate the double integral $\iint\limits_R {\left( {x - y} \right)dxdy}.$ The region of integration $$R$$ is bounded

$x = 0, x = 1, y = x,y = 2 - {x^2}.$

### Example 2

Calculate the integral $\iint\limits_R {\left( {x + y} \right)dxdy}.$ The region of integration $$R$$ is bounded by the lines

$x = 0, y = 0, x + y = 2.$

### Example 3

Evaluate the integral $\iint\limits_R {xdxdy},$ where the region of integration $$R$$ is bounded by the graphs of the functions

$y = {x^3}, x + y = 2, x = 0.$

### Example 4

Find the integral $\iint\limits_R {{x^2}ydxdy},$ where the region $$R$$ is the segment of a circle. The boundaries of the segment are defined by the equations

${x^2} + {y^2} = 4, x + y - 2 = 0.$

### Example 1.

Calculate the double integral $\iint\limits_R {\left( {x - y} \right)dxdy}.$ The region of integration $$R$$ is bounded

$x = 0, x = 1, y = x,y = 2 - {x^2}.$

Solution.

We can represent the region $$R$$ in the form $${R =}$$ $${ \left\{ {\left( {x,y} \right)|\;0 \le x \le 1,\;}\right.}$$ $${\left.{ x \le y \le 2 - {x^2}} \right\}.}$$

$R = \left\{ {\left( {x,y} \right)|\;0 \le x \le 1,\; x \le y \le 2 - {x^2}} \right\}.$

So $$R$$ is the region of type $$I$$ (see Figure $$1$$).

According to the Fubini's formula,

$\iint\limits_R {\left( {x - y} \right)dxdy} = \int\limits_0^1 {\int\limits_x^{2 - {x^2}} {\left( {x - y} \right)dydx} } = \int\limits_0^1 {\left[ {\int\limits_x^{2 - {x^2}} {\left( {x - y} \right)dy} } \right]dx} .$

Calculate first the inner integral:

$\int\limits_x^{2 - {x^2}} {\left( {x - y} \right)dy} = \left. {\left( {xy - \frac{{{y^2}}}{2}} \right)} \right|_{y = x}^{2 - {x^2}} = [ {x\left( {2 - {x^2}} \right) - \frac{{{{\left( {2 - {x^2}} \right)}^2}}}{2}}] - \left[ {{x^2} - \frac{{{x^2}}}{2}} \right] = - \frac{{{x^4}}}{2} - {x^3} + \frac{{3{x^2}}}{2} + 2x - 2.$

Now we can compute the outer integral:

$\int\limits_0^1 {\left( { - \frac{{{x^4}}}{2} - {x^3} + \frac{{3{x^2}}}{2} + 2x - 2} \right)dx} = \left. {\left( { - \frac{{{x^5}}}{{10}} - \frac{{{x^4}}}{4} + \frac{{{x^3}}}{2} + {x^2} - 2x} \right)} \right|_0^1 = - \frac{{17}}{{20}}.$

### Example 2.

Calculate the integral $\iint\limits_R {\left( {x + y} \right)dxdy}.$ The region of integration $$R$$ is bounded by the lines

$x = 0, y = 0, x + y = 2.$

Solution.

We can represent the region $$R$$ as the set

$R = \left\{ {\left( {x,y} \right)|\;0 \le x \le 2,\; 0 \le y \le 2 - x} \right\}.$

The region $$R$$ belongs to type $$I.$$ Converting the double integral into the iterated one, we get

$\iint\limits_R {\left( {x + y} \right)dxdy} = \int\limits_0^2 {\int\limits_0^{2 - x} {\left( {x + y} \right)dydx} } = \int\limits_0^2 {\left[ {\int\limits_0^{2 - x} {\left( {x + y} \right)dy} } \right]dx} = \int\limits_0^2 {\left[ {\left. {\left( {xy + \frac{{{y^2}}}{2}} \right)} \right|_{y = 0}^{2 - x}} \right]dx} = \int\limits_0^2 {\left[ {x\left( {2 - x} \right) + \frac{{{{\left( {2 - x} \right)}^2}}}{2}} \right]dx} = \int\limits_0^2 {\left( {2 - \frac{{{x^2}}}{2}} \right)dx} = \left. {\left( {2x - \frac{{{x^3}}}{6}} \right)} \right|_0^2 = \frac{8}{3}.$

### Example 3.

Evaluate the integral $\iint\limits_R {xdxdy},$ where the region of integration $$R$$ is bounded by the graphs of the functions

$y = {x^3}, x + y = 2, x = 0.$

Solution.

The region of integration $$R$$ is shown in Figure $$3.$$

The curve $$y = {x^3}$$ and the line $$x + y = 2$$ intersect at $$\left( {1,1} \right).$$ Then, by Fubini's theorem, the double integral is

$\iint\limits_R {xdxdy} = \int\limits_0^1 {\int\limits_{{x^3}}^{2 - x} {xdydx} } = \int\limits_0^1 {\left[ {\int\limits_{{x^3}}^{2 - x} {xdy} } \right]dx} = \int\limits_0^1 {\left[ {\left. {\left( {xy} \right)} \right|_{y = {x^3}}^{2 - x}} \right]dx} = \int\limits_0^1 {\left[ {x\left( {2 - x} \right) - {x^4}} \right]dx} = \left. {\left( {{x^2} - \frac{{{x^3}}}{3} - \frac{{{x^5}}}{5}} \right)} \right|_0^1 = 1 - \frac{1}{3} - \frac{1}{5} = \frac{7}{{15}}.$

### Example 4.

Find the integral $\iint\limits_R {{x^2}ydxdy},$ where the region $$R$$ is the segment of a circle. The boundaries of the segment are defined by the equations

${x^2} + {y^2} = 4, x + y - 2 = 0.$

Solution.

The circle $${x^2} + {y^2} = 4$$ has the radius $$2$$ and centre at the origin (Figure $$4$$).

Since the upper half of the circle is equivalent to $$y = \sqrt {4 - {x^2}},$$ the double integral can be written in the following form:
$\iint\limits_R {{x^2}ydxdy} = \int\limits_0^2 {\int\limits_{2 - x}^{\sqrt {4 - {x^2}} } {{x^2}ydxdy} } = \int\limits_0^2 {\left[ {\int\limits_{2 - x}^{\sqrt {4 - {x^2}} } {{x^2}ydy} } \right]dx} = \int\limits_0^2 {\left[ {\left. {\left( {\frac{{{x^2}{y^2}}}{2}} \right)} \right|_{y = 2 - x}^{\sqrt {4 - {x^2}} }} \right]dx} = \frac{1}{2}\int\limits_0^2 {\left[ {{x^2}\left( {4 - {x^2}} \right) - {x^2}{{\left( {2 - x} \right)}^2}} \right]dx} = \frac{1}{2}\int\limits_0^2 {\left[ {4{x^2} - {x^4} - {x^2}\left( {4 - 4x + {x^2}} \right)} \right]dx} = \frac{1}{2}\int\limits_0^2 {\left( {4{x^3} - 2{x^4}} \right)dx} = \frac{1}{2}\left. {\left( {{x^4} - \frac{{2{x^5}}}{5}} \right)} \right|_0^2 = \frac{1}{2}\left( {16 - \frac{{64}}{5}} \right) = \frac{8}{5}.$