Calculus

Double Integrals

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Geometric Applications of Double Integrals

  • 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

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

    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

    Click a problem to see the 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 \(x = 1,\) \(x = 2.\)

    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

    \[{{x_1} = 0,\;\;}\kern-0.3pt{{y_1} = a + 0 = a,}\]

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

    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

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

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