Calculus

Double Integrals

Double Integrals Logo

Double Integrals over Rectangular Regions

  • Let \(R\) be a rectangular region\(\left[ {a,b} \right] \times \left[ {c,d} \right]\) of the \(xy\)-plane. Then using the Fubini’s theorem we can write the double integral in this region through the iterated integral:

    \[
    {\iint\limits_R {f\left( {x,y} \right)dxdy} }
    = {\int\limits_a^b {\left( {\int\limits_c^d {f\left( {x,y} \right)dy} } \right)dx} }
    = {\int\limits_c^d {\left( {\int\limits_a^b {f\left( {x,y} \right)dx} } \right)dy} .}
    \]

    The region \(R\) here is simultaneously the region of type \(I\) and type \(II,\) so that we have a free choice as to whether to integrate \(f\left( {x,y} \right)\) with respect to \(x\) or \(y\) first. It is usually better to evaluate the easier integral first.

    In the special case where the integrand \(f\left( {x,y} \right)\) can be written as the product of two functions \(g\left( {x} \right) h\left( {y} \right),\) we have

    \[
    {\iint\limits_R {f\left( {x,y} \right)dxdy} }
    = {\iint\limits_R {g\left( x \right)h\left( y \right)dxdy} }
    = {\left( {\int\limits_a^b {g\left( x \right)dx} } \right)\cdot}\kern0pt{ \left( {\int\limits_c^d {h\left( y \right)dy} } \right)}
    \]


    Solved Problems

    Click a problem to see the solution.

    Example 1

    Evaluate the double integral \(\iint\limits_R {xydxdy}\) over the rectangular region \({R =}\) \({ \left\{ {\left( {x,y} \right)|\;2 \le x \le 5,\; }\right.}\) \({\left.{ 0 \le y \le 1} \right\}.}\)

    Example 2

    Calculate the double integral \(\iint\limits_R {x{y^2}dxdy}\) over the region \({R =}\) \({ \left\{ {\left( {x,y} \right)|\;1 \le x \le 5,\; }\right.}\) \({\left.{ 0 \le y \le 2} \right\}.}\)

    Example 3

    Calculate the integral \(\iint\limits_R {\large\frac{{dxdy}}{{{{\left( {x + y} \right)}^2}}}\normalsize}\) over the rectangular region \({R =}\) \({ \left\{ {\left( {x,y} \right)|\;0 \le x \le 2,\; }\right.}\) \({\left.{ 1 \le y \le 2} \right\}.}\)

    Example 4

    Evaluate the integral \(\iint\limits_R {\cos \left( {x + y} \right)dxdy} \) over the region \({R =}\) \({ \left\{ {\left( {x,y} \right)|\;0 \le x \le {\large\frac{\pi }{4}\normalsize},\;}\right.}\) \({\left.{ 0 \le y \le {\large\frac{\pi }{4}\normalsize}} \right\}.}\)

    Example 5

    Evaluate the integral \(\iint\limits_R {\left( {x – {y^2}} \right)dxdy}\) over the region \({R =}\) \({ \left\{ {\left( {x,y} \right)|\;2 \le x \le 3,\;}\right.}\) \({\left.{ 1 \le y \le 2} \right\}.}\)

    Example 1.

    Evaluate the double integral \(\iint\limits_R {xydxdy}\) over the rectangular region \({R =}\) \({ \left\{ {\left( {x,y} \right)|\;2 \le x \le 5,\; }\right.}\) \({\left.{ 0 \le y \le 1} \right\}.}\)

    Solution.

    We see that the integrand \(f\left( {x,y} \right)\) is the product \(g\left( {x} \right) h\left( {y} \right).\) Then we have

    \[
    {\iint\limits_R {xydxdy} = \int\limits_2^4 {xdx} \cdot \int\limits_0^1 {ydy} }
    = {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_2^4 \cdot \left. {\left( {\frac{{{y^2}}}{2}} \right)} \right|_0^1 }
    = {\left( {8 – 2} \right)\left( {\frac{1}{2} – 0} \right) }={ 3.}
    \]

    Example 2.

    Calculate the double integral \(\iint\limits_R {x{y^2}dxdy}\) over the region \({R =}\) \({ \left\{ {\left( {x,y} \right)|\;1 \le x \le 5,\; }\right.}\) \({\left.{ 0 \le y \le 2} \right\}.}\)

    Solution.

    Since the integrand \(f\left( {x,y} \right)\) is the product \(g\left( {x} \right) h\left( {y} \right),\) we can write

    \[
    {\iint\limits_R {x{y^2}dxdy} }
    = {\int\limits_1^5 {xdx} \cdot \int\limits_0^2 {{y^2}dy} }
    = {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_1^5 \cdot \left. {\left( {\frac{{{y^3}}}{3}} \right)} \right|_0^2 }
    = {\left( {\frac{{25}}{2} – \frac{1}{2}} \right)\left( {\frac{8}{3} – 0} \right) }={ 64.}
    \]

    Page 1
    Problems 1-2
    Page 2
    Problems 3-5