Write the integral in the form
\\[ \\int_{\\simplify[fractionNumbers]{{t1script()}pi}}^{d\\pi} \\int_a^b f(r,\\theta) \\, \\mathrm{d}r \\, \\mathrm{d}\\theta .\\]
$a=\\,$[[0]] $b=\\,$[[1]]
\n$d=\\,$[[2]]
\n$f(r,\\theta)=\\,$[[3]](fully expanded)
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\nAs we move anticlockwise around the origin $\\theta$ increases, so the upper limit is $\\simplify[fractionNumbers]{{t2script()}pi}$.
\nAs for $f(r,\\theta)$, the first term changes to
\n\\[ \\simplify{{a}x^{m}} = \\simplify{{a}r^{m}}\\cos^\\var{m}\\theta \\]
\nThe 2nd and 3rd terms can be rearranged so that
\n\\begin{align}
\\simplify{{b}x^2} + \\simplify{{c}y^2}
&= \\simplify{{b}(x^2 + y^2)} &+& \\simplify{{c-b}y^2} \\\\
&= \\simplify{{b}r^2} &+& \\simplify{{c-b}r^2}\\sin^2{\\theta},
\\end{align}
The last term is a constant, so it doesn't change.
\nFinally, when we change an integral from cartesian to polar form we need to multiply the function by $r$.
\nSo the whole definite integral in polar form is
\n\\[ \\int_\\simplify[fractionNumbers]{{t1script()}pi}^\\simplify[fractionNumbers]{{t2script()}pi} \\int_{\\var{r1}}^{\\var{r2}} \\simplify{{a}r^{m+1}}\\cos^\\var{m}{\\theta} + \\simplify{{c-b}r^3}\\sin^2{\\theta}+ \\simplify{{b}r^3} + \\var{d}r \\, \\mathrm{d}r \\, \\mathrm{d}\\theta .\\]
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"}, "tags": [], "name": "2. Change of Form (Cartesian to Polar)", "ungrouped_variables": [], "statement": "Convert this integral from cartesian to polar form, and define the limits of integration from the domain.
\n\\[ \\iint \\simplify{{a}x^{m}} + \\simplify{{b}x^2} + \\simplify{{c}y^2} + \\var{d} \\, \\mathrm{d}y \\, \\mathrm{d}x \\]
\n{domain()}
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