19 lines
1.2 KiB
Markdown
19 lines
1.2 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 |