Find the area of the region in the xy-plane given by $0\leq y \leq 4x, 9\leq49x^2−y^2 \leq11$ Find the area of the region in the xy-plane given by $0\leq y \leq 4x, 9\leq49x^2−y^2 \leq11$ 
how to start this problem i really don't get any idea for this problem what are limits for x and y 
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\bracks{P} = 1\ \mbox{whenever}\ P\ \mbox{is}\ true\ \mbox{and, otherwise, equal to}\ 0}$. $\ds{\bracks{P}}$ is an Iverson Bracket. They are very convenient when we are dealing with 'region definitions'.

\begin{align}
&\iint_{\large\mathbb{R}^{2}}\bracks{0 \leq y \leq 4x}
\bracks{9 \leq 49x^{2} - y^{2} \leq 11}\dd x\,\dd y
\\[5mm] = &\
{1 \over 7}\int_{0}^{\infty}\int_{0}^{\infty}\bracks{y \leq {4 \over 7}\,x}
\bracks{9 \leq x^{2} - y^{2} \leq 11}\dd x\,\dd y
\end{align}

With Cylindrical
Coordinates $\ds{x = r\cos\pars{\phi}\,,\ y = r\sin\pars{\phi}}$
$\ds{\pars{~\mbox{with}\ r > 0\,,\ \phi \in \left[0,2\pi\right)~}}$:

\begin{align}
&\iint_{\large\mathbb{R}^{2}}\bracks{0 \leq y \leq 4x}
\bracks{9 \leq 49x^{2} - y^{2} \leq 11}\dd x\,\dd y
\\[5mm] = &\
{1 \over 7}\int_{0}^{\pi/2}\int_{0}^{\infty}
\bracks{\sin\pars{\phi} \leq {4 \over 7}\,\cos\pars{\phi}}
\bracks{-11 \leq r^{2}\cos\pars{2\phi} \leq -9}r\,\dd r\,\dd\phi
\\[5mm] = &\
{1 \over 14}\int_{0}^{\pi/2}\int_{0}^{\infty}
\bracks{{1 - \cos\pars{2\phi} \over 2} \leq
{16 \over 49}\,{1 + \cos\pars{2\phi} \over 2}}
\bracks{-11 \leq r\cos\pars{2\phi} \leq -9}\,\dd r\,\dd\phi
\\[5mm] = &\
{1 \over 28}\int_{0}^{\pi}\int_{0}^{\infty}
\bracks{1 - \cos\pars{\phi} \leq
{16 \over 49}\,\braces{1 + \cos\pars{\phi}}}
\bracks{-11 \leq r\cos\pars{\phi} \leq -9}\,\dd r\,\dd\phi
\\[5mm] = &\
{1 \over 28}\int_{0}^{\pi/2}\int_{0}^{\infty}
\bracks{1 + \sin\pars{\phi} \leq
{16 \over 49}\,\braces{1 - \sin\pars{\phi}}}
\bracks{-11 \leq -r\sin\pars{\phi} \leq -9}\,\dd r\,\dd\phi
\\[5mm] = &\
{1 \over 28}\int_{0}^{\pi/2}\int_{0}^{\infty}
\bracks{\sin\pars{\phi} \geq {33 \over 65}}
\bracks{9 \leq r\sin\pars{\phi} \leq 11}\,\dd r\,\dd\phi =
{1 \over 28}\int_{\phi_{0}}^{\pi/2}\int_{9/\sin\pars{\phi}}^{11/\sin\pars{\phi}}
\,\dd r\,\dd\phi
\end{align}

where $\ds{\phi_{0} = \arcsin\pars{33 \over 65}}$.


Then,
\begin{align}
&\iint_{\large\mathbb{R}^{2}}\bracks{0 \leq y \leq 4x}
\bracks{9 \leq 49x^{2} - y^{2} \leq 11}\dd x\,\dd y =
{1 \over 14}\int_{\phi_{0}}^{\pi/2}\csc\pars{\phi}\,\dd\phi
\\[5mm] = &\
\bbx{\ds{{1 \over 14}\,\ln\pars{11 \over 3}}} \approx 0.0928
\end{align}
