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Double Integrals

# Double Integrals over General Regions

Page 1
Problems 1-2
Page 2
Problems 3-9

If the region $$R$$ of type $$I$$ is bounded by $$x = a,$$ $$x = b,$$ $$y = p\left( x \right)$$ and $$y = q\left( x \right)$$ with $$a \lt b$$ and $$p\left( x \right) \lt q\left( x \right)$$ for all $$x \in \left[ {a,b} \right],$$ 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 on problem description to see solution.

### ✓Example 1

Calculate the double integral $$\iint\limits_R {\left( {x – y} \right)dxdy}.$$ The region of integration $$R$$ is bounded by $$x = 0,$$ $$x = 1,$$ $$y = x,$$ $$y = 2 – {x^2}.$$

### ✓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 5

Find the double integral $$\iint\limits_R {\sin \left( {x + y} \right)dxdy},$$ defined in the region $$R$$ bounded by the lines $$y = x,$$ $$x + y = {\large\frac{\pi }{2}\normalsize},$$ $$y = 0.$$

### ✓Example 6

Find the integral $$\iint\limits_R {ydydx},$$ where $$R$$ is bounded by the straight line $$y = 2x$$ and parabola $$y = 3 – {x^2}.$$

### ✓Example 7

Calculate the double integral $$\iint\limits_R {x\sin ydydx},$$ where $$R$$ is bounded by $$y = 0,$$ $$y = {x^2},$$ $$x = 1.$$

### ✓Example 8

Calculate the integral $$\iint\limits_R {{e^x}dxdy}.$$ The region of integration is the triangle with the vertices $$O\left( {0,0} \right),$$ $$B\left( {0,1} \right),$$ and $$C\left( {1,1} \right).$$

### ✓Example 9

Find the double integral $$\iint\limits_R {\left( {x + y} \right)dxdy},$$ where the region $$R$$ is a parallelogram with the sides $$y = x,$$ $$y = x + a,$$ $$y = a,$$ $$y = 2a,$$ $$a$$ is a parameter.

### Example 1.

Calculate the double integral $$\iint\limits_R {\left( {x – y} \right)dxdy}.$$ The region of integration $$R$$ is bounded by $$x = 0,$$ $$x = 1,$$ $$y = x,$$ $$y = 2 – {x^2}.$$

#### Solution.

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

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:

Figure 1.

${\int\limits_x^{2 – {x^2}} {\left( {x – y} \right)dy} } = {\left. {\left( {xy – \frac{{{y^2}}}{2}} \right)} \right|_{y = x}^{2 – {x^2}} } = {[ {x\left( {2 – {x^2}} \right) }}-{{ \frac{{{{\left( {2 – {x^2}} \right)}^2}}}{2}}] }-{ \left[ {{x^2} – \frac{{{x^2}}}{2}} \right] } = { – \frac{{{x^4}}}{2} – {x^3} + \frac{{3{x^2}}}{2} }+{ 2x – 2.}$ Now we can compute the outer integral:
${\int\limits_0^1 {\left( { – \frac{{{x^4}}}{2} – {x^3} }\right.}+{\left.{ \frac{{3{x^2}}}{2} + 2x – 2} \right)dx} } = {\left. {\left( { – \frac{{{x^5}}}{{10}} – \frac{{{x^4}}}{4} }\right.}\kern0pt{+ \left.{ \frac{{{x^3}}}{2} + {x^2} – 2x} \right)} \right|_0^1 } = { – \frac{{17}}{{20}}.}$

### Example 2.

Calculate the integral $$\iint\limits_R {\left( {x + y} \right)dxdy}.$$ The region of integration $$R$$ is bounded by the lines $$x = 0,$$ $$y = 0,$$ $$x + y = 2.$$

Figure 2.

#### Solution.

We can represent the region $$R$$ as the set $${R =}$$ $${ \left\{ {\left( {x,y} \right)|\;0 \le x \le 2,\;}\right.}$$ $${\left.{ 0 \le y \le 2 – x} \right\}}$$ (Figure $$2$$). The region $$R$$ belongs to type $$I.$$ Converting the double integral into the iterated one, we get
${\iint\limits_R {\left( {x + y} \right)dxdy} } = {\int\limits_0^2 {\int\limits_0^{2 – x} {\left( {x + y} \right)dydx} } } = {\int\limits_0^2 {\left[ {\int\limits_0^{2 – x} {\left( {x + y} \right)dy} } \right]dx} }$

$= {\int\limits_0^2 {\left[ {\left. {\left( {xy + \frac{{{y^2}}}{2}} \right)} \right|_{y = 0}^{2 – x}} \right]dx} } = {\int\limits_0^2 {\left[ {x\left( {2 – x} \right) + \frac{{{{\left( {2 – x} \right)}^2}}}{2}} \right]dx} }\kern0pt = {\int\limits_0^2 {\left( {2 – \frac{{{x^2}}}{2}} \right)dx} }={ \left. {\left( {2x – \frac{{{x^3}}}{6}} \right)} \right|_0^2 }={ \frac{8}{3}.}$

Page 1
Problems 1-2
Page 2
Problems 3-9