How do I find area under a circle between specific x values

Question

Im sorry if this is a dumb question because I'm not very familiar with calculus.

I need to find the area under the circle $x^2+y^2=9$ between x = $√5$ and $-√(5)$

I cant find a way to do this even with u-substitution

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\iint_{\mathbb{R}^{2}}\bracks{x^{2} + y^{2} < 9} \bracks{-\root{5} < x < \root{5}}\dd^{2}\vec{r}} \\[5mm] = & \ \int_{0}^{2\pi}\int_{0}^{3}\bracks{r\verts{\cos\pars{\theta}} < \root{5}} r\,\dd r\,\dd\theta \\[5mm] = & \ \int_{0}^{3}r\int_{0}^{2\pi}\bracks{\verts{\cos\pars{\theta}}< {\root{5} \over r}}\dd\theta\,\dd r \\[5mm] = & \ \int_{0}^{3}r\int_{-\pi}^{\pi} \bracks{\verts{\cos\pars{\theta}} < {\root{5} \over r}}\dd\theta\,\dd r \\[5mm] & \ 2\int_{0}^{3}r\int_{0}^{\pi}\bracks{\verts{\cos\pars{\theta}} < {\root{5} \over r}}\dd\theta\,\dd r \\[5mm] = & \ 2\int_{0}^{3}r\int_{-\pi/2}^{\pi/2} \bracks{\verts{\sin\pars{\theta}} < {\root{5} \over r}} \dd\theta\,\dd r \\[5mm] = & \ 4\int_{0}^{3}r\int_{0}^{\pi/2}\bracks{\sin\pars{\theta} < {\root{5} \over r}}\dd\theta\,\dd r \\[5mm] = & \ 4\int_{0}^{3}r\bracks{\bracks{r > \root{5}} \int_{0}^{\arcsin\pars{\root{5}/r}}\dd\theta + \bracks{r < \root{5}}\int_{0}^{\pi/2}\dd\theta}\dd r \\[5mm] = & \ 4\int_{\root{5}}^{3}r\arcsin\pars{\root{5} \over r}\dd r + 4\int_{0}^{\root{5}}r\pars{\pi \over 2}\dd r \\[5mm] = & \ \bbx{\color{#44f}{4\root{5} + 18\on{arccsc}\pars{3\root{5} \over 5}}} \approx 24.0835 \\ & \end{align}

Answer 2

Observe that the area in question, by symmetry, is the same as multiplying $\int_0^{\sqrt{5}}9-x^2dx$ by $4$. This integral is about $6.02$, so the answer is approximately $24.08$