TeachingCalculus
- Multiple integrals
- Double integrals over rectangles
- Double integrals over general regions
- Double integrals in polar coordinates
- Applications of double integrals
- Triple integrals
- Triple integrals in cylindrical coordinates
- Triple integrals in spherical coordinates
- Change of variables in multiple integrals
- More applications 1
- More applications 2
Calculus – Multiple integrals – Double integrals over general regions
Suppose that \(D\) is a bounded region, which can be enclosed in a rectangular region \(R\):
Now we define a new function \(F\) with domain \(R\) by:
\[F(x,y)=\left\{\begin{array}{ll}f(x,y),&(x,y)\in D\\[2.5mm]0,&(x,y)\notin D.\end{array}\right.\]If \(f\) is integrable over \(R\), then we define the integral of \(f\) over \(D\) by:
\[\iint\limits_Df(x,y)\,dA=\iint\limits_RF(x,y)\,dA.\] |
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Regions of type I
A region \(D\) in \(\mathbb{R}^2\) is called of type I if it lies between the graphs of two continuous functions of \(x\), that is:
\[D=\{(x,y)\,|\,a\leq x\leq b,\;g_1(x)\leq y\leq g_2(x)\},\]where \(g_1\) and \(g_2\) are continuous functions on the interval \([a,b]\).
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If \(f\) is continuous on a region \(D\) of type I, that is
then
\[\iint\limits_Df(x,y)\,dA=\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy\,dx.\]Regions of type II
A region \(D\) in \(\mathbb{R}^2\) is called of type II if it lies between the graphs of two continuous functions of \(y\), that is:
where \(h_1\) and \(h_2\) are continuous functions on the interval \([c,d]\).
If \(f\) is continuous on a region \(D\) of type II, that is
then
\[\iint\limits_Df(x,y)\,dA=\int_c^d\int_{h_1(y)}^{h_2(y)}f(x,y)\,dx\,dy.\]More general regions
If \(D=D_1\cup D_2\), then
\(D\) is neither of type I, nor of type II.
\(D=D_1\cup D_2\), where \(D_1\) is of type I and \(D_2\) is of type II.
Examples
1) Stewart §15.2, Example 1
Evaluate \(\displaystyle\iint\limits_D(x+2y)\,dA\), where \(D\) is the region bounded by the parabolas \(y=2x^2\) and \(y=1+x^2\).
Solution: The parabolas intersect when \(2x^2=1+x^2\) or equivalently \(x^2=1\), so when \(x=\pm1\). Note that the region \(D\) is of type I and not of type II:
\[D=\{(x,y)\,|\,-1\leq x\leq 1,\;2x^2\leq y\leq 1+x^2\}.\]Hence we have:
\begin{align*} \iint\limits_D(x+2y)\,dA&=\int_{-1}^1\int_{2x^2}^{1+x^2}(x+2y)\,dy\,dx=\int_{-1}^1\bigg[xy+y^2\bigg]_{y=2x^2}^{1+x^2}\,dx =\int_{-1}^1\left(x(1+x^2)+(1+x^2)^2-x\cdot2x^2-(2x^2)^2\right)\,dx\\[2.5mm] &=\int_{-1}^1\left(1+x+2x^2-x^3-3x^4\right)\,dx=\bigg[x+\frac{1}{2}x^2+\frac{2}{3}x^3-\frac{1}{4}x^4-\frac{3}{5}x^5\bigg]_{-1}^1 =2\left(1+\frac{2}{3}-\frac{3}{5}\right)=\frac{32}{15}. \end{align*}2) Stewart §15.2, Example 2
Find the volume of the solid that lies under the paraboloid \(z=x^2+y^2\) and above the region \(D\) in the \(xy\)-plane bounded by the line
\(y=2x\) and the parabola \(y=x^2\)..
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Solution: Note that the region \(D\) is both of type I and of type II. As a type I region we have:
\[D=\{(x,y)\,|\,0\leq x\leq 2,\;x^2\leq y\leq 2x\}.\]Therefore the volume under \(z=x^2+y^2\) and above \(D\) is
\begin{align*} V&=\iint\limits_D(x^2+y^2)\,dA=\int_0^2\int_{x^2}^{2x}(x^2+y^2)\,dy\,dx=\int_0^2\bigg[x^2y+\frac{1}{3}y^3\bigg]_{y=x^2}^{2x}\,dx =\int_0^2\left(x^2\cdot2x+\frac{1}{3}(2x)^3-x^2\cdot x^2-\frac{1}{3}(x^2)^3\right)\,dx\\[2.5mm] &=\int_0^2\left(\frac{14}{3}x^3-x^4-\frac{1}{33}x^6\right)\,dx=\bigg[\frac{7}{6}x^4-\frac{1}{5}x^5-\frac{1}{21}x^7\bigg]_{x=0}^2 =\frac{56}{3}-\frac{32}{5}-\frac{128}{21}=\frac{216}{35}. \end{align*}As a type II region we have:
\[D=\{(x,y)\,|\,0\leq y\leq 4,\;\textstyle\frac{1}{2}y\leq x\leq\sqrt{y}\}.\]Therefore the volume can also be obtained as
\begin{align*} V&=\iint\limits_D(x^2+y^2)\,dA=\int_0^4\int_{\frac{1}{2}y}^{\sqrt{y}}(x^2+y^2)\,dx\,dy=\int_0^4\bigg[\frac{1}{3}x^3+xy^2\bigg]_{x=\frac{1}{2}y}^{\sqrt{y}}\,dy =\int_0^4\left(\frac{1}{3}(\sqrt{y})^3+y^2\sqrt{y}-\frac{1}{24}y^3-\frac{1}{2}y^3\right)\,dy\\[2.5mm] &=\int_0^4\left(\frac{1}{3}y\sqrt{y}+y^2\sqrt{y}-\frac{13}{24}y^3\right)\,dy=\bigg[\frac{2}{15}y^2\sqrt{y}+\frac{2}{7}y^3\sqrt{y}-\frac{13}{96}y^4\bigg]_{y=0}^4 =\frac{64}{15}+\frac{256}{7}-\frac{104}{3}=\frac{216}{35}. \end{align*}3) Stewart §15.2, Example 3
Evaluate \(\displaystyle\iint\limits_Dxy\,dA\), where \(D\) is the region bounded by the line \(y=x-1\) and the parabola \(y^2=2x+6\).
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Solution: Again the region \(D\) is both of type I and of type II. As a type I region we have to split the integral into two parts:
\[\iint\limits_Dxy\,dA=\int_{-3}^{-1}\int_{-\sqrt{2x+6}}^{\sqrt{2x+6}}xy\,dy\,dx+\int_{-1}^5\int_{x-1}^{\sqrt{2x+6}}xy\,dy\,dx.\]As a type II region it is much easier:
\[D=\{(x,y)\,|\,-2\leq y\leq 4,\;\textstyle\frac{1}{2}y^2-3\leq x\leq y+1\}.\]Therefore we obtain
\begin{align*} \iint\limits_Dxy\,dA&=\int_{-2}^4\int_{\frac{1}{2}y^2-3}^{y+1}xy\,dx\,dy=\int_{-2}^4\bigg[\frac{1}{2}x^2y\bigg]_{\frac{1}{2}y^2-3}^{y+1}\,dy =\frac{1}{2}\int_{-2}^4y\left((y+1)^2-\left(\textstyle\frac{1}{2}y^2-3\right)^2\right)\,dy\\[2.5mm] &=\frac{1}{2}\int_{-2}^4\left(-8y+2y^2+4y^3-\frac{1}{4}y^5\right)\,dy =\frac{1}{2}\bigg[-4y^2+\frac{2}{3}y^3+y^4-\frac{1}{24}y^6\bigg]_{y=-2}^4\\[2.5mm] &=-32+\frac{64}{3}+128-\frac{256}{3}+8+\frac{8}{3}-8+\frac{4}{3}=36. \end{align*}4) Stewart §15.2, Example 4
Find the volume of the tetrahedron bounded by the planes \(x+2y+z=2\), \(x=2y\), \(x=0\) and \(z=0\).
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Solution: Note that, since the plane \(x+2y+z=2\) intersects the \(xy\)-plane (\(z=0\)) in the line \(x+2y=2\), the volume of the tetrahedron \(T\) is the volume under the graph of the function \(z=2-x-2y\) above the triangular region \(D\) given by
\[D=\{(x,y)\,|\,0\leq x\leq 1,\;\textstyle\frac{1}{2}x\leq y\leq1-\frac{1}{2}x\}.\]Therefore the volume equals
\begin{align*} V&=\iint\limits_D(2-x-2y)\,dA=\int_0^1\int_{\frac{1}{2}x}^{1-\frac{1}{2}x}(2-x-2y)\,dy\,dx=\int_0^1\bigg[2y-xy-y^2\bigg]_{y=\frac{1}{2}x}^{1-\frac{1}{2}x}\,dx\\[2.5mm] &=\int_0^1\left(2-x-x\left(1-\frac{1}{2}x\right)-\left(1-\frac{1}{2}x\right)^2-x+\frac{1}{2}x^2+\frac{1}{4}x^2\right)\,dx =\int_0^1(1-2x+x^2)\,dx=\frac{1}{3}(x-1)^3\bigg|_0^1=\frac{1}{3}. \end{align*}5) Stewart §15.2, Example 5
Evaluate the iterated integral \(\displaystyle\int_0^1\int_x^1\sin(y^2)\,dy\,dx\).
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Solution: The inner integral \(\displaystyle\int\sin(y^2)\,dy\) cannot be evaluated in terms of elementary functions. However, note that
\[\int_0^1\int_x^1\sin(y^2)\,dy\,dx=\iint\limits_D\sin(y^2)\,dA,\]where
\[D=\{(x,y)\,|\,0\leq x\leq 1,\;x\leq y\leq 1\}.\]This region \(D\) is both of type I and of type II. As a type II region it is given by
\[D=\{(x,y)\,|\,0\leq y\leq 1,\;0\leq x\leq y\}.\]Hence we have
\[\int_0^1\int_x^1\sin(y^2)\,dy\,dx=\iint\limits_D\sin(y^2)\,dA=\int_0^1\int_0^y\sin(y^2)\,dx\,dy=\int_0^1y\sin(y^2)\,dy =\bigg[-\frac{1}{2}\cos(y^2)\bigg]_{y=0}^1=\frac{1}{2}\left(1-\cos(1)\right).\]The area of a region
If we integrate the constant function \(f(x,y)=1\) over a region \(D\), we get the area of \(D\): \(\displaystyle\textrm{area}(D)=\iint\limits_D1\,dA\).
Corollary: If \(m\leq f(x,y)\leq M\) for all \((x,y)\in D\), then: \(\displaystyle m\cdot\textrm{area}(D)\leq\iint\limits_Df(x,y)\,dA\leq M\cdot\textrm{area}(D)\).
Example: Consider the region \(D\) in \(\mathbb{R}^2\) bounded by the graphs of \(y=x^2\) and \(y=\sqrt{x}\).
Then we have:
\[\textrm{area}(D)=\iint\limits_D1\,dA=\int_0^1\int_{x^2}^{\sqrt{x}}1\,dy\,dx=\int_0^1\left(\sqrt{x}-x^2\right)\,dx =\bigg[\frac{2}{3}x\sqrt{x}-\frac{1}{3}x^3\bigg]_{x=0}^1=\frac{1}{3}.\]Last modified on September 27, 2021
