Calculus

Double Integrals

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Double Integrals over General Regions

  • If a region \(R\) of type \(I\) is bounded by \(x = a,\) \(x = b,\) \(y = p\left( x \right)\) and \(y = q\left( x \right)\) with \(a \lt b\) and \(p\left( x \right) \lt q\left( x \right)\) for all \(x \in \left[ {a,b} \right],\) then using the Fubini’s theorem the double integral can be written as

    \[
    {\iint\limits_R {f\left( {x,y} \right)dxdy} }
    = {\int\limits_{x = a}^{x = b} {\int\limits_{y = p\left( x \right)}^{y = q\left( x \right)} {f\left( {x,y} \right)dydx} } .}
    \]

    The similar formula exists for regions of type \(II.\) If a region \(R\) of type \(II\) is bounded by the graphs of the functions \(x = u\left( y \right),\) \(x = v\left( y \right),\) \(y = c,\) \(y = d\) provided that \(c \lt d\) and \(u\left( y \right) \lt v\left( y \right)\) for all \(y \in \left[ {c,d} \right],\) then the double integral over the region \(R\) is expressed through the iterated integral by the Fubini’s theorem

    \[
    {\iint\limits_R {f\left( {x,y} \right)dxdy} }
    = {\int\limits_{y = c}^{y = d} {\int\limits_{x = u\left( y \right)}^{x = v\left( y \right)} {f\left( {x,y} \right)dxdy} } .}
    \]

    It is sometimes useful to break the region \(R\) up into two or more smaller regions, and integrate over each separately.


    Solved Problems

    Click a problem to see the solution.

    Example 1

    Calculate the double integral \(\iint\limits_R {\left( {x – y} \right)dxdy}.\) The region of integration \(R\) is bounded by \(x = 0,\) \(x = 1,\) \(y = x,\) \(y = 2 – {x^2}.\)

    Example 2

    Calculate the integral \(\iint\limits_R {\left( {x + y} \right)dxdy}.\) The region of integration \(R\) is bounded by the lines \(x = 0,\) \(y = 0,\) \(x + y = 2.\)

    Example 3

    Evaluate the integral \(\iint\limits_R {xdxdy},\) where the region of integration \(R\) is bounded by the graphs of the functions \(y = {x^3},\) \(x + y = 2,\) \(x = 0.\)

    Example 4

    Find the integral \(\iint\limits_R {{x^2}ydxdy},\) where the region \(R\) is the segment of a circle. The boundaries of the segment are defined by the equations \({x^2} + {y^2} = 4,\) \(x + y – 2 = 0.\)

    Example 5

    Find the double integral \(\iint\limits_R {\sin \left( {x + y} \right)dxdy},\) defined in the region \(R\) bounded by the lines \(y = x,\) \(x + y = {\large\frac{\pi }{2}\normalsize},\) \(y = 0.\)

    Example 6

    Find the integral \(\iint\limits_R {ydydx},\) where \(R\) is bounded by the straight line \(y = 2x\) and parabola \(y = 3 – {x^2}.\)

    Example 7

    Calculate the double integral \(\iint\limits_R {x\sin ydydx},\) where \(R\) is bounded by \(y = 0,\) \(y = {x^2},\) \(x = 1.\)

    Example 8

    Calculate the integral \(\iint\limits_R {{e^x}dxdy}.\) The region of integration is the triangle with the vertices \(O\left( {0,0} \right),\) \(B\left( {0,1} \right),\) and \(C\left( {1,1} \right).\)

    Example 9

    Find the double integral \(\iint\limits_R {\left( {x + y} \right)dxdy},\) where the region \(R\) is a parallelogram with the sides \(y = x,\) \(y = x + a,\) \(y = a,\) \(y = 2a,\) \(a\) is a parameter.

    Example 1.

    Calculate the double integral \(\iint\limits_R {\left( {x – y} \right)dxdy}.\) The region of integration \(R\) is bounded by \(x = 0,\) \(x = 1,\) \(y = x,\) \(y = 2 – {x^2}.\)

    Solution.

    We can represent the region \(R\) in the form \({R =}\) \({ \left\{ {\left( {x,y} \right)|\;0 \le x \le 1,\;}\right.}\) \({\left.{ x \le y \le 2 – {x^2}} \right\}.}\)

    So \(R\) is the region of type \(I\) (see Figure \(1\)).

    A region of integration bounded by straight lines and a parabola
    Figure 1.

    According to the Fubini’s formula,

    \[
    {\iint\limits_R {\left( {x – y} \right)dxdy} }
    = {\int\limits_0^1 {\int\limits_x^{2 – {x^2}} {\left( {x – y} \right)dydx} } }
    = {\int\limits_0^1 {\left[ {\int\limits_x^{2 – {x^2}} {\left( {x – y} \right)dy} } \right]dx} .}
    \]

    Calculate first the inner integral:

    \[
    {\int\limits_x^{2 – {x^2}} {\left( {x – y} \right)dy} }
    = {\left. {\left( {xy – \frac{{{y^2}}}{2}} \right)} \right|_{y = x}^{2 – {x^2}} }
    = {[ {x\left( {2 – {x^2}} \right) }}-{{ \frac{{{{\left( {2 – {x^2}} \right)}^2}}}{2}}] }-{ \left[ {{x^2} – \frac{{{x^2}}}{2}} \right] }
    = { – \frac{{{x^4}}}{2} – {x^3} + \frac{{3{x^2}}}{2} }+{ 2x – 2.}
    \]

    Now we can compute the outer integral:

    \[
    {\int\limits_0^1 {\left( { – \frac{{{x^4}}}{2} – {x^3} }\right.}+{\left.{ \frac{{3{x^2}}}{2} + 2x – 2} \right)dx} }
    = {\left. {\left( { – \frac{{{x^5}}}{{10}} – \frac{{{x^4}}}{4} }\right.}\kern0pt{+ \left.{ \frac{{{x^3}}}{2} + {x^2} – 2x} \right)} \right|_0^1 }
    = { – \frac{{17}}{{20}}.}
    \]

    Example 2.

    Calculate the integral \(\iint\limits_R {\left( {x + y} \right)dxdy}.\) The region of integration \(R\) is bounded by the lines \(x = 0,\) \(y = 0,\) \(x + y = 2.\)

    Solution.

    We can represent the region \(R\) as the set \({R =}\) \({ \left\{ {\left( {x,y} \right)|\;0 \le x \le 2,\;}\right.}\) \({\left.{ 0 \le y \le 2 – x} \right\}}\) (Figure \(2\)).

    A region of integration bounded by a straight line and coordinate axes
    Figure 2.

    The region \(R\) belongs to type \(I.\) Converting the double integral into the iterated one, we get

    \[ {\iint\limits_R {\left( {x + y} \right)dxdy} } = {\int\limits_0^2 {\int\limits_0^{2 – x} {\left( {x + y} \right)dydx} } } = {\int\limits_0^2 {\left[ {\int\limits_0^{2 – x} {\left( {x + y} \right)dy} } \right]dx} }={\int\limits_0^2 {\left[ {\left. {\left( {xy + \frac{{{y^2}}}{2}} \right)} \right|_{y = 0}^{2 – x}} \right]dx} } = {\int\limits_0^2 {\left[ {x\left( {2 – x} \right) + \frac{{{{\left( {2 – x} \right)}^2}}}{2}} \right]dx} }\kern0pt = {\int\limits_0^2 {\left( {2 – \frac{{{x^2}}}{2}} \right)dx} }={ \left. {\left( {2x – \frac{{{x^3}}}{6}} \right)} \right|_0^2 }={ \frac{8}{3}.} \]

    Page 1
    Problems 1-2
    Page 2
    Problems 3-9