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

@@ -16,4 +16,23 @@ Find the volume of the solid below $z=1-x^2-y^2$ and above the first quadrant on
 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}$$