Calculus

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

A type I region of integration

Figure 1.

A type II region of integration

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

A region of integration in polar coordinates

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 \(x = 1,\) \(x = 2.\)

Region of integration bounded by two hyperbolas and two vertical lines

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

Region of integration bounded by a parabola and a straight line

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