Answer:
Step-by-step explanation:
[tex]s=y, t = y- x^2[/tex]
[tex]x^2=y-t[/tex]
[tex]x=\sqrt{s-t}[/tex]
[tex]y=0, y=16, y=x^2, and\\ y=x^2-5.[/tex]
[tex]y-x^2=0,y-x^2=-5[/tex]
therefore,
[tex]s=[(s,t)10\leq s\leq 16,-3\leq t \leq 0][/tex]
Taking partial derivatives
[tex]\frac{dx}{ds} =\frac{1}{2\sqrt{s-t} } \\\\\frac{dx}{dt} =\frac{1}{2\sqrt{s-t} } \\\\\frac{dy}{ds} =1\\\\\frac{dy}{dt} =0[/tex]
[tex]\frac{\delta (x,y)}{\delta (s,t)} =\left|\begin{array}{ccc}\frac{1}{2\sqrt{s-t} } &-\frac{1}{2\sqrt{s-t} } \\1&0\end{array}\right|[/tex]
[tex]=\frac{1}{2\sqrt{s-t} }[/tex]
[tex]\int\limits \int\limits_R {x} \, dx \, dy=\int\limits\int\limits_s {x} \sqrt{s-t} |\frac{\delta(x,y)}{\delta (s,t)} |dsdt[/tex]
[tex]=\int\limits^{16}_0\int\limits^0_{-5} \sqrt{s-t} \frac{1}{2\sqrt{s-t} } dt ds[/tex]
[tex]=\frac{1}{2} \int\limits^{16}_0 ds\int\limits^0_{-5} \, dt \\\\=\frac{1}{2} [s]^{16}_0[t]^0_{-5}\\\\=\frac{1}{2} (16)(5)\\\\=40[/tex]
