vault backup: 2026-10-05 15:06:34

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ben committed 2026-10-05 15:06:34 -07:00
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+5 -7
@@ -17,9 +17,6 @@ $$ Area(x^*)=\int_{x}^{x^2}x^*ydy$$
Since $x$ is bound by $1<x<2$ the double integral to find the volume can be written as Since $x$ is bound by $1<x<2$ the double integral to find the volume can be written as
$$\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \ dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \ \left.\frac{xy^2}{2} \right\vert_{x}^{x^2} \ \right] dx \\[1.5ex] \int_{1}^2 \left[ \frac{x}{2}((x^2)^2-(x)^2) \right] dx \\[1.5ex] \int_{1}^2 \frac{x^5}{2}-\frac{x^3}{2} dx \\[1.5ex] \left.\frac{x^6}{12}-\frac{x^4}{8} \right\vert_{1}^2 \\[1.5ex] \left( \frac{16}{3} - 2 \right) - \left( \frac{1}{12} - \frac{1}{8} \right) \\[1.5ex] \frac{27}{8} \end{gather}$$ $$\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \ dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \ \left.\frac{xy^2}{2} \right\vert_{x}^{x^2} \ \right] dx \\[1.5ex] \int_{1}^2 \left[ \frac{x}{2}((x^2)^2-(x)^2) \right] dx \\[1.5ex] \int_{1}^2 \frac{x^5}{2}-\frac{x^3}{2} dx \\[1.5ex] \left.\frac{x^6}{12}-\frac{x^4}{8} \right\vert_{1}^2 \\[1.5ex] \left( \frac{16}{3} - 2 \right) - \left( \frac{1}{12} - \frac{1}{8} \right) \\[1.5ex] \frac{27}{8} \end{gather}$$
> [!NOTE] Theorem
> Contents
### Theorem: ### Theorem:
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 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
$$\int \int_{D} f(x,y)$$ $$\int \int_{D} f(x,y)$$
@@ -27,7 +24,8 @@ can be rewritten as
$$\int_{a}^b \int_{g_{1}(x)}^{g_{2}(x)} f(x,y) \ dydx$$ $$\int_{a}^b \int_{g_{1}(x)}^{g_{2}(x)} f(x,y) \ dydx$$
This can also be done if $D$ is bound by functions on the x-axis instead of the y-axis This can also be done if $D$ is bound by functions on the x-axis instead of the y-axis
### Property:
If $D=D_{1} \cup D_{2}$ and $D_{1}$ and $D_{2}$ do not intersect except at their boundaries then > [!note] Property
$$\int \int_{D} f(x,y) \ dA = \int \int_{D_{1}} f(x,y) \ dA + \int \int_{D_{2}} f(x,y) \ dA$$ > If $D=D_{1} \cup D_{2}$ and $D_{1}$ and $D_{2}$ do not intersect except at their boundaries then
This can be used to split up integrals similar to area in 2D > $$\int \int_{D} f(x,y) \ dA = \int \int_{D_{1}} f(x,y) \ dA + \int \int_{D_{2}} f(x,y) \ dA$$
> This can be used to split up integrals similar to area in 2D