Convert ${}_{R}f(x,y)dA$ into polar coordinates, where $f(x,y)=3x+y$ and $R$ is the region in the first quadrant bounded by $y=\sqrt[\phantom{2}]{1-x^2},y=\sqrt[\phantom{2}]{36-x^2},y=0,$ and $y=\sqrt[\phantom{2}]{3}x.$

$drd$

$={\displaystyle \underset{\frac{}{4}}{\overset{\frac{}{2}}{}}}{\displaystyle \underset{3}{\overset{5}{}}}{r}^3$ubrace$((co{s}^2+sincos)ubrace)drd$

$={\displaystyle \underset{\frac{}{4}}{\overset{\frac{}{4}}{}}}\frac{1}{4}{r}^4(co{s}^2)(lncos){|}_r||=3^r=5dr$

$={\displaystyle \underset{\frac{}{4}}{\overset{\frac{}{2}}{}}}(\frac{1}{4}{5}^4-\frac{1}{4}{3}^4)(co{s}^2+sincos)d$

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Vouititbe

Finally, the value of your double integral is $$