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College/AU 26/MATH 224 (Multivar)/Chapter 15.3 - Double Integrals in Polar Coordinates.md

@@ -11,28 +11,37 @@ if a function is defined as $f(x,y)$ and the region it is integrated over is pol
 $$\int_{\alpha}^\beta \int_{a}^b f(r\cos(\theta), r\sin(\theta))r \ dr d\theta$$
 
 
-Find the volume of the solid below $z=1-x^2-y^2$ and above the first quadrant on the xy plane
+> [!example] Example
+> Find the volume of the solid below $z=1-x^2-y^2$ and above the first quadrant on the xy plane
+> 
+> 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 $\sqrt{ 1-z }$
+> 
+> since the slice of the equation at $z=0$ forms the circle $x^2+y^2=1$ the bounds 
+> $$\begin{gather}
+> 	0\leq r \leq 1 \\ 0 \leq \theta \leq \frac{\pi}{2}
+> \end{gather}$$
+> can be used to find the volume of the solid.
+> 
+> writing out the integral with these bounds gives
+> $$\begin{gather} 
+>  \int \int_{R} f(x,y) dA \\[2ex] \int_{\alpha}^\beta \int_{a}^b f(r\cos(\theta), r\sin(\theta))r \ dr d\theta \\[2ex]
+> 	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-(r\cos \theta)^2-(r\sin \theta)^2)r \ dr d\theta
+> \end{gather}$$
+> This integral can now be solved similar to other double integrals
+> $$\begin{gather}
+> 	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-(r\cos \theta)^2-(r\sin \theta)^2)r \ dr d\theta  \\[2ex]
+> 	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-r^2\cos^2 \theta-r^2\sin^2 \theta)r \ dr d\theta \\[2ex] 
+> 	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-r^2)r \ dr d\theta \\[2ex]
+> 	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 r-r^3 \ dr d\theta \\[2ex]
+> 	\int_{0}^{\frac{\pi}{2}} \left( \left.\frac{r^2}{2}-\frac{r^4}{4} \right\vert_{0}^1 \right) d\theta \\[2ex] 
+> 	\int_{0}^{\frac{\pi}{2}} \frac{1}{4} d\theta \\[2ex] 
+> 	V=\frac{\pi}{8}
+> \end{gather}$$
 
-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 $\sqrt{ 1-z }$
 
-since the slice of the equation at $z=0$ forms the circle $x^2+y^2=1$ the bounds 
+When are polar coordinates useful for evaluating double integrals?
-$$\begin{gather}
+- Domain has a simple formula in terms of $(r,\theta)$
-	0\leq r \leq 1 \\ 0 \leq \theta \leq \frac{\pi}{2}
+- Integrand has a simple formula in terms of $(r, \theta)$
-\end{gather}$$
-can be used to find the volume of the solid.
 
-writing out the integral with these bounds gives
+Polar regions with functions as boundaries can also be integrated similar to cartesian ones
-$$\begin{gather} 
+$$\int_{\alpha}^\beta \int_{h_{1}(\theta)}^{h_{2}(\theta)} f(r\cos \theta, r\sin \theta)r \ dr d\theta$$
- \int \int_{R} f(x,y) dA \\[2ex] \int_{\alpha}^\beta \int_{a}^b f(r\cos(\theta), r\sin(\theta))r \ dr d\theta \\[2ex]
-	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-(r\cos \theta)^2-(r\sin \theta)^2)r \ dr d\theta
-\end{gather}$$
-This integral can now be solved similar to other double integrals
-$$\begin{gather}
-	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-(r\cos \theta)^2-(r\sin \theta)^2)r \ dr d\theta  \\[2ex]
-	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-r^2\cos^2 \theta-r^2\sin^2 \theta)r \ dr d\theta \\[2ex] 
-	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 (1-r^2)r \ dr d\theta \\[2ex]
-	\int_{0}^{\frac{\pi}{2}} \int_{0}^1 r-r^3 \ dr d\theta \\[2ex]
-	\int_{0}^{\frac{\pi}{2}} \left( \left.\frac{r^2}{2}-\frac{r^4}{4} \right\vert_{0}^1 \right) d\theta \\[2ex] 
-	\int_{0}^{\frac{\pi}{2}} \frac{1}{4} d\theta \\[2ex] 
-	V=\frac{\pi}{8}
-\end{gather}$$