30 lines
1.8 KiB
Markdown
30 lines
1.8 KiB
Markdown
#uw/notes #uw/class/math224
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- - -
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Recall that:
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$$\int \int_{R} f(x, y) dA$$
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is the signed area under the function $f(x,y)$ over the domain $R$ where
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$$R=[a,b] \times [c,d]$$
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and it can be written as
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$$\int_{a}^b \int_{c}^d f(x, y) dy dx$$
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This can be extended to more general domains that are not rectangles such as $[a,b] \times [c,d]$
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Example:
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finding the volume under the function $f(x,y) = xy$ over the domain bound by
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$$\begin{gathered} 1<x<2 \\[1.5ex] x < y < x^2 \end{gathered}$$
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We can think of finding the volume of the solid by using slices parallel to the $YZ$ plane. With the domain provided this area can be found with
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$$ Area(x^*)=\int_{x}^{x^2}x^*ydy$$
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Since $x$ is bound by $1<x<2$ the double integral to find the volume can be written as
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$$\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \ dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \ \left.\frac{xy^2}{2} \right\vert_{x}^{x^2} \ \right] dx \\[1.5ex] \int_{1}^2 \left[ \frac{x}{2}((x^2)^2-(x)^2) \right] dx \\[1.5ex] \int_{1}^2 \frac{x^5}{2}-\frac{x^3}{2} dx \\[1.5ex] \left.\frac{x^6}{12}-\frac{x^4}{8} \right\vert_{1}^2 \\[1.5ex] \left( \frac{16}{3} - 2 \right) - \left( \frac{1}{12} - \frac{1}{8} \right) \end{gather}$$
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27/8
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Theorem:
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If the region on the $XY$ plane for the region being integrated over of $f(x,y)$ ($D$) is continuous and bound by $x=a$ and $x=b$ in the x-axis and two functions $g_{1}(x)$ and $g_{2}(x)$ on the y-axis then the integral
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$$\int \int_{D} f(x,y)$$
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can be rewritten as
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$$\int_{a}^b \int_{g_{1}(x)}^{g_{2}(2)} f(x,y) \ dydx$$
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This can also be done if $D$ is bound by functions on the x-axis instead of the y-axis
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If $D=D_{1} D_{2}$ and $D_{1}$ and $D_{2}$ do not intersect except at their boundaries then
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$$\int \int_{D} f(x,y)dA = \int \int_{D_{1}} f(x,y) \ dA + \int \int_{D_{2}} f(x,y)$$ |