# Geometric Applications of Double Integrals

## Areas

If f (x, y) = 1 in the integral $$\iint\limits_R {f\left( {x,y} \right)dxdy},$$ then the double integral gives the area of the region R.

The area of a type I region (Figure 1) can be written in the form:

$A = \int\limits_a^b {\int\limits_{g\left( x \right)}^{h\left( x \right)} {dydx} } .$

Similarly, the area of a type $$II$$ region (Figure $$2$$) is given by the formula

$A = \int\limits_c^d {\int\limits_{p\left( y \right)}^{q\left( y \right)} {dxdy} } .$

## Volumes

If $${f\left( {x,y} \right)} \gt 0$$ over a region $$R,$$ then the volume of the solid below the surface $$z = {f\left( {x,y} \right)}$$ and above $$R$$ is expressed as

$V = \iint\limits_R {f\left( {x,y} \right)dA}.$

If $$R$$ is a type $$I$$ region bounded by $$x = a,$$ $$x = b,$$ $$y = g\left( x \right),$$ $$y = h\left( x \right),$$ the volume of the solid is

$V = \iint\limits_R {f\left( {x,y} \right)dA} = \int\limits_a^b {\int\limits_{g\left( x \right)}^{h\left( x \right)} {f\left( {x,y} \right)dydx} } .$

Similarly, if $$R$$ is a type $$II$$ region bounded by $$y = c,$$ $$y = d,$$ $$x = p\left( y \right),$$ $$x = q\left( y \right),$$ the volume of the solid is given by

$V = \iint\limits_R {f\left( {x,y} \right)dA} = \int\limits_c^d {\int\limits_{p\left( y \right)}^{q\left( y \right)} {f\left( {x,y} \right)dxdy} } .$

If $$f\left( {x,y} \right) \ge g\left( {x,y} \right)$$ over a region $$R,$$ then the volume of the cylindrical solid between the surfaces $${z_1} = g\left( {x,y} \right)$$ and $${z_2} = f\left( {x,y} \right)$$ over $$R$$ is given by

$V = \iint\limits_R {\left[ {f\left( {x,y} \right) - g\left( {x,y} \right)} \right]dA}.$

## Surface Area

We assume that the surface is given as a graph of function $$z = g\left( {x,y} \right),$$ and the domain of this function is a region $$R.$$ Then the area of the surface over the region $$R$$ is

$S = \iint\limits_R {\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy} ,$

provided that the derivatives $${\frac{{\partial z}}{{\partial x}}}$$ and $${\frac{{\partial z}}{{\partial y}}}$$ are continuous over the region $$R.$$

## Areas and Volumes in Polar Coordinates

If $$S$$ is a region in the $$xy$$-plane bounded by $$\theta = \alpha,$$ $$\theta = \beta,$$ $$r = h\left( \theta \right),$$ $$r = g\left( \theta \right)$$ (Figure $$3$$), then the area of the region is defined by the formula

$A = \iint\limits_R {dA} = \int\limits_\alpha ^\beta {\int\limits_{h\left( \theta \right)}^{g\left( \theta \right)} {rdrd\theta } } .$

The volume of the solid below $$z = f\left( {r,\theta } \right)$$ over a region $$S$$ in polar coordinates is given by

$V = \iint\limits_S {f\left( {r,\theta } \right)rdrd\theta } .$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the area of the region $$R$$ bounded by the hyperbolas $y = {\frac{{{a^2}}}{x}}, y = {\frac{{2{a^2}}}{x}} \left( {a \gt 0} \right)$ and the vertical lines $$x = 1,$$ $$x = 2.$$

### Example 2

Find the area of the region $$R$$ bounded by ${y^2} = {a^2} - ax, y = a + x.$

### Example 3

Find the volume of the solid in the first octant bounded by the planes $y = 0, z = 0, z = x, z + x = 4.$

### Example 4

Describe the solid whose volume is given by the integral $V = \int\limits_0^1 {dx} \int\limits_0^{1 - x} {\left( {{x^2} + {y^2}} \right)dy}.$

### Example 1.

Find the area of the region $$R$$ bounded by the hyperbolas $y = {\frac{{{a^2}}}{x}}, y = {\frac{{2{a^2}}}{x}} \left( {a \gt 0} \right)$ and the vertical lines $$x = 1,$$ $$x = 2.$$

Solution.

The region $$R$$ is sketched in Figure $$4.$$

Using the formula for the area of a type $$I$$ region

$A = \iint\limits_R {dxdy} = \int\limits_a^b {\int\limits_{g\left( x \right)}^{h\left( x \right)} {dydx} },$

we have

$A = \iint\limits_R {dxdy} = \int\limits_1^2 {\left[ {\int\limits_{\frac{{{a^2}}}{x}}^{\frac{{2{a^2}}}{x}} {dy} } \right]dx} = \int\limits_1^2 {\left[ {\left. y \right|_{\frac{{{a^2}}}{x}}^{\frac{{2{a^2}}}{x}}} \right]dx} = \int\limits_1^2 {\left( {\frac{{2{a^2}}}{x} - \frac{{{a^2}}}{x}} \right)dx} = {a^2}\int\limits_1^2 {\frac{{dx}}{x}} = {a^2}\left( {\ln 2 - \ln 1} \right) = {a^2}\ln 2.$

### Example 2.

Find the area of the region $$R$$ bounded by ${y^2} = {a^2} - ax, y = a + x.$

Solution.

We first determine the points of intersection of the two curves.

$\left\{ \begin{array}{l} {y^2} = {a^2} - ax\\ y = a + x \end{array} \right.,\;\; \Rightarrow \left( {a + x} \right)^2 = {a^2} - ax,\;\; \Rightarrow {a^2} + 2ax + {x^2} = {a^2} - ax,\;\; \Rightarrow {x^2} + 3ax = 0,\;\; \Rightarrow x\left( {x + 3a} \right) = 0,\;\; \Rightarrow {x_{1,2}} = 0;\; - 3a.$

So the coordinates of the points of intersection are

${x_1} = 0,\;\; {y_1} = a + 0 = a,$
${x_2} = - 3a,\;\; {y_2} = a - 3a = - 2a.$

It is simpler to consider $$R$$ as a type $$II$$ region $$\left({\text{Figure }5}\right).$$

To calculate the area of the region, we transform the equations of the boundaries:

${y^2} = {a^2} - ax,\;\; \Rightarrow ax = {a^2} - {y^2},\;\; \Rightarrow x = a - \frac{{{y^2}}}{a},$
$y = a + x,\;\; \Rightarrow x = y - a.$

Then we have

$A = \iint\limits_R {dxdy} = \int\limits_{ - 2a}^a {\left[ {\int\limits_{y - a}^{a - \frac{{{y^2}}}{a}} {dx} } \right]dy} = \int\limits_{ - 2a}^a {\left[ {\int\limits_{y - a}^{a - \frac{{{y^2}}}{a}} {dx} } \right]dy} = \int\limits_{ - 2a}^a {\left[ {\left. x \right|_{y - a}^{a - \frac{{{y^2}}}{a}}} \right]dy} = \int\limits_{ - 2a}^a {\left[ {a - \frac{{{y^2}}}{a} - \left( {y - a} \right)} \right]dy} = \int\limits_{ - 2a}^a {\left( {2a - \frac{{{y^2}}}{a} - y} \right)dy} = \left. {\left( {2ay - \frac{{{y^3}}}{{3a}} - \frac{{{y^2}}}{2}} \right)} \right|_{ - 2a}^a = \left( {2{a^2} - \frac{{{a^3}}}{{3a}} - \frac{{{a^2}}}{2}} \right) - \left( { - 4{a^2} + \frac{{8{a^3}}}{{3a}} - \frac{{4{a^2}}}{2}} \right) = \frac{{9{a^2}}}{2}.$

### Example 3.

Find the volume of the solid in the first octant bounded by the planes $y = 0, z = 0, z = x, z + x = 4.$

Solution.

The given solid is shown in Figure $$6.$$

As it can be seen from the figure, the base $$R$$ is the square in the first quadrant. For given $$x$$ and $$y,$$ the $$z$$-value in the solid varies from $$z = x$$ to $$z = 4 - x.$$ Then the volume is

$V = \iint\limits_R {\left[ {\left( {4 - x} \right) - x} \right]dxdy} = \int\limits_0^2 {\left[ {\int\limits_0^2 {\left( {4 - 2x} \right)dy} } \right]dx} = \int\limits_0^2 {\left[ {\left. {\left( {4y - 2xy} \right)} \right|_{y = 0}^2} \right]dx} = \int\limits_0^2 {\left( {8 - 4x} \right)dx} = \left. {\left( {8x - 2{x^2}} \right)} \right|_0^2 = 16 - 8 = 8.$

### Example 4.

Describe the solid whose volume is given by the integral $V = \int\limits_0^1 {dx} \int\limits_0^{1 - x} {\left( {{x^2} + {y^2}} \right)dy}.$

Solution.

The given solid (Figures $$7,8$$) lies above the triangle $$R$$ in the $$xy$$-plane, bounded by the coordinate axes $$Ox,$$ $$Oy$$ and the straight line $$y = 1 - x,$$ and under the paraboloid $$z = {x^2} + {y^2}.$$

The volume of the solid is

$V = \int\limits_0^1 {dx} \int\limits_0^{1 - x} {\left( {{x^2} + {y^2}} \right)dy} = \int\limits_0^1 {\left[ {\left. {\left( {{x^2}y + \frac{{{y^3}}}{3}} \right)} \right|_{y = 0}^{1 - x}} \right]dx} = \int\limits_0^1 {\left[ {{x^2}\left( {1 - x} \right) + \frac{{{{\left( {1 - x} \right)}^3}}}{3}} \right]dx} = \int\limits_0^1 {\left( {2{x^2} - \frac{{4{x^3}}}{3} - x + \frac{1}{3}} \right)dx} = \left. {\left( {\frac{{2{x^3}}}{3} - \frac{4}{3} \cdot \frac{{{x^4}}}{4} - \frac{{{x^2}}}{2} + \frac{x}{3}} \right)} \right|_0^1 = \frac{2}{3} - \frac{1}{3} - \frac{1}{2} + \frac{1}{3} = \frac{1}{6}.$

See more problems on Page 2.