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

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

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

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

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

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

A region of integration in polar coordinates — Figure 3.

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

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 \(4.\)

Region of integration bounded by two hyperbolas and two vertical lines — 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 \(\left({\text{Figure }5}\right).\)

Region of integration bounded by a parabola and a straight line — Figure 5.

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 \(z = {x^2} + {y^2}.\)

A solid over a triangle under a paraboloid — Figure 7.

A triangular region of integration — Figure 8.

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.