38 lines
1.8 KiB
Markdown
38 lines
1.8 KiB
Markdown
#uw/notes #uw/class/math224
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Instead of being bound by a rectangular region regions in polar coordinates are often bound with
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$$\begin{gather}
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a\leq r\leq b \\ \alpha \leq \theta \leq \beta
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\end{gather}$$
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![[PolarRegion.excalidraw]]
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if a function is defined as $f(x,y)$ and the region it is integrated over is polar the integral will often look similar to
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$$\int_{\alpha}^\beta \int_{a}^b f(r\cos(\theta), r\sin(\theta))r \ dr d\theta$$
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Find the volume of the solid below $z=1-x^2-y^2$ and above the first quadrant on the xy plane
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rearranging the equation gives $x^2+y^2=1-z$ which shows that each horizontal slice of the function is a circle centered at $(0,0)$ with a radius of $\sqrt{ 1-z }$
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since the slice of the equation at $z=0$ forms the circle $x^2+y^2=1$ the bounds
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$$\begin{gather}
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0\leq r \leq 1 \\ 0 \leq \theta \leq \frac{\pi}{2}
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\end{gather}$$
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can be used to find the volume of the solid.
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writing out the integral with these bounds gives
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$$\begin{gather}
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\int \int_{R} f(x,y) dA \\[2ex] \int_{\alpha}^\beta \int_{a}^b f(r\cos(\theta), r\sin(\theta))r \ dr d\theta \\[2ex]
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\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-(r\cos \theta)^2-(r\sin \theta)^2)r \ dr d\theta
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\end{gather}$$
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This integral can now be solved similar to other double integrals
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$$\begin{gather}
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\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-(r\cos \theta)^2-(r\sin \theta)^2)r \ dr d\theta \\[2ex]
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\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-r^2\cos^2 \theta-r^2\sin^2 \theta)r \ dr d\theta \\[2ex]
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\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-r^2)r \ dr d\theta \\[2ex]
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\int_{0}^{\frac{\pi}{2}} \int_{0}^1 r-r^3 \ dr d\theta \\[2ex]
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\int_{0}^{\frac{\pi}{2}} \left( \left.\frac{r^2}{2}-\frac{r^4}{4} \right\vert_{0}^1 \right) d\theta \\[2ex]
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\int_{0}^{\frac{\pi}{2}} \frac{1}{4} d\theta \\[2ex]
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V=\frac{\pi}{8}
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\end{gather}$$ |