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Double Integrals

Geometric Applications of Double Integrals

Page 1
Problems 1-2
Page 2
Problems 3-10

Areas

If $$f\left( {x,y} \right) = 1$$ 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 $$1\text{)}$$ 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
$A = \int\limits_c^d {\int\limits_{p\left( y \right)}^{q\left( y \right)} {dxdy} } .$

Figure 1.

Figure 2.

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
${S \text{=}}\kern-0.3pt{\iint\limits_R {\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy} }$ provided that the derivatives $${\large\frac{{\partial z}}{{\partial x}}\normalsize}$$ and $${\large\frac{{\partial z}}{{\partial y}}\normalsize}$$ 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
${V} ={ \iint\limits_S {f\left( {r,\theta } \right)rdrd\theta } .}$

Figure 3.

Solved Problems

Click on problem description to see solution.

✓Example 1

Find the area of the region $$R$$ bounded by the hyperbolas $$y = {\large\frac{{{a^2}}}{x}\normalsize},$$ $$y = {\large\frac{{2{a^2}}}{x}\normalsize}$$ $$\left( {a \gt 0} \right)$$ 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 5

Find the volume of the solid bounded by $$z = xy,$$ $$x + y = a,$$ $$z = 0.$$

✓Example 6

Find the volume of the solid bounded by the surfaces $$z = 0,$$ $$x + y = 1,$$ $${x^2} + {y^2} = 1,$$ $$z = 1 – x.$$

✓Example 7

Find the area of one loop of the rose defined by the equation $$r = \cos 2\theta.$$

✓Example 8

Find the volume of the unit sphere.

✓Example 9

Use polar coordinates to find the volume of a right circular cone with height $$H$$ and a circular base with radius $$R$$ (see Figure $$15$$).

✓Example 10

Calculate the surface area of a sphere of radius $$a.$$

Example 1.

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

Figure 4.

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
${{x_1} = 0,\;\;}\kern-0.3pt{{y_1} = a + 0 = a,}$ ${{x_2} = – 3a,\;\;}\kern-0.3pt{{y_2} = a – 3a = – 2a.}$

Figure 5.

The given region $$R$$ is shown in Figure $$5.$$ It is simpler to consider $$R$$ as a type $$II$$ region. 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
${A = \iint\limits_R {dxdy} } = {\int\limits_{ – 2a}^a {\left[ {\int\limits_{y – a}^{a – \frac{{{y^2}}}{a}} {dx} } \right]dy} } = {\int\limits_{ – 2a}^a {\left[ {\int\limits_{y – a}^{a – \frac{{{y^2}}}{a}} {dx} } \right]dy} } = {\int\limits_{ – 2a}^a {\left[ {\left. x \right|_{y – a}^{a – \frac{{{y^2}}}{a}}} \right]dy} } = {\int\limits_{ – 2a}^a {\left[ {a – \frac{{{y^2}}}{a} – \left( {y – a} \right)} \right]dy} } = {\int\limits_{ – 2a}^a {\left( {2a – \frac{{{y^2}}}{a} – y} \right)dy} } = {\left. {\left( {2ay – \frac{{{y^3}}}{{3a}} – \frac{{{y^2}}}{2}} \right)} \right|_{ – 2a}^a } = {\left( {2{a^2} – \frac{{{a^3}}}{{3a}} – \frac{{{a^2}}}{2}} \right) }-{ \left( { – 4{a^2} + \frac{{8{a^3}}}{{3a}} – \frac{{4{a^2}}}{2}} \right) } = {\frac{{9{a^2}}}{2}.}$

Page 1
Problems 1-2
Page 2
Problems 3-10