Prelazimo u polarne koordinate
\[
x=\rho\cos\varphi,\qquad y=\rho\sin\varphi,\qquad dx\,dy=\rho\,d\rho\,d\varphi.
\]
U polarnim koordinatama uslovi za skup \(D\) postaju
\[
x\leq x^{2}+y^{2}\leq2x\quad\Longleftrightarrow\quad\rho\cos\varphi\leq\rho^{2}\leq2\rho\cos\varphi.
\]
Kako je \(\rho\ge0\), deljenjem sa \(\rho\) dobijamo
\[
\cos\varphi\leq\rho\leq2\cos\varphi.
\]
Da bi interval za \(\rho\) bio nenegativan i nenulti, mora biti \(\cos\varphi\ge0\),
tj. \(\varphi\in[-\frac{\pi}{2},\frac{\pi}{2}]\).
Uslov \(y\leq x\sqrt{3}\) u polarnim koordinatama daje
\[
\rho\sin\varphi\leq\rho\cos\varphi\sqrt{3}\quad\Longrightarrow\quad\tan\varphi\leq\sqrt{3}.
\]
Na intervalu \([-\frac{\pi}{2},\frac{\pi}{2}]\) ovo znači \(\varphi\leq\frac{\pi}{3}\).
Dakle \(\varphi\in\bigl(-\frac{\pi}{2},\frac{\pi}{3}\bigr].\)
Za ove \(\varphi\) važi \(arctg\frac{y}{x}=arctg(\tan\varphi)=\varphi.\)
Dakle dobijamo
\[
I=\iint\limits_{D} arctg\frac{y}{x}\,dxdy=\int_{-\pi/2}^{\pi/3}\int_{\cos\varphi}^{2\cos\varphi}\varphi\cdot\rho\,d\rho\,d\varphi.
\]
Prvo računamo unutrašnji integral po \(\rho\):
\[
\int_{\cos\varphi}^{2\cos\varphi}\rho\,d\rho=\frac{1}{2}\bigl((2\cos\varphi)^{2}-(\cos\varphi)^{2}\bigr)=\frac{3}{2}\cos^{2}\varphi.
\]
Koristimo identitet \(\cos^{2}\varphi=\frac{1+\cos2\varphi}{2}\), dobijamo
\[
I=\frac{3}{2}\int_{-\pi/2}^{\pi/3}\varphi\cdot\frac{1+\cos2\varphi}{2}\,d\varphi=\frac{3}{4}\left(\int_{-\pi/2}^{\pi/3}\varphi\,d\varphi+\int_{-\pi/2}^{\pi/3}\varphi\cos2\varphi\,d\varphi\right).
\]
Prvi integral je
\[
\int_{-\pi/2}^{\pi/3}\varphi\,d\varphi=\frac{1}{2}\Bigl(\bigl(\frac{\pi}{3}\bigr)^{2}-\bigl(-\frac{\pi}{2}\bigr)^{2}\Bigr)=\frac{1}{2}\Bigl(\frac{\pi^{2}}{9}-\frac{\pi^{2}}{4}\Bigr)=-\frac{5\pi^{2}}{72}.
\]
Drugi rešavamo parcijalnom integracijom sa \(u=\varphi,\; dv=\cos2\varphi\,d\varphi\):
\[
\int\varphi\cos2\varphi\,d\varphi=\frac{1}{2}\varphi\sin2\varphi+\frac{1}{4}\cos2\varphi+C,
\]
pa je
\[
\int_{-\pi/2}^{\pi/3}\varphi\cos2\varphi\,d\varphi=\frac{1}{2}\Bigl(\frac{\pi}{3}\sin\frac{2\pi}{3}-\bigl(-\frac{\pi}{2}\bigr)\sin(-\pi)\Bigr)+\frac{1}{4}\Bigl(\cos\frac{2\pi}{3}-\cos(-\pi)\Bigr).
\]
Menjamo vrednosti: \(\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2},\;\sin(-\pi)=0,\;\cos\frac{2\pi}{3}=-\frac{1}{2},\;\cos(-\pi)=-1.\)
Dakle
\[
\int_{-\pi/2}^{\pi/3}\varphi\cos2\varphi\,d\varphi=\frac{1}{2}\cdot\frac{\pi}{3}\cdot\frac{\sqrt{3}}{2}+\frac{1}{4}\Bigl(-\frac{1}{2}-(-1)\Bigr)=\frac{\pi\sqrt{3}}{12}+\frac{1}{8}.
\]
Na kraju
\[
I=\frac{3}{4}\Bigl(-\frac{5\pi^{2}}{72}+\frac{\pi\sqrt{3}}{12}+\frac{1}{8}\Bigr)=-\frac{5\pi^{2}}{96}+\frac{\pi\sqrt{3}}{16}+\frac{3}{32}.
\]
3. Izračunati masu tela određenog relacijama
\[9\leq x^{2}+y^{2}+z^{2}\leq 16,\quad z\geq0,\quad y\leq0,\]
ako je gustina data sa \(\rho(x,y,z)=3(x^{2}+y^{2})\).
Iz uslova \(9\le x^{2}+y^{2}+z^{2}\le16\) sledi \(3\le\rho\le4.\)
Iz \(z\ge0\) dobijamo \(\cos\theta\ge0\), tj. \(\theta\in[0,\frac{\pi}{2}].\)
Uslov \(y\le0\) u sfernim koordinatama glasi \(\sin\varphi\le0\), pa je \(\varphi\in[\pi,2\pi].\)
Gustina u sfernim koordinatama je
\[
\rho(x,y,z)=3(x^{2}+y^{2})
=3\bigl(\rho^{2}\sin^{2}\theta(\cos^{2}\varphi+\sin^{2}\varphi)\bigr)
=3\rho^{2}\sin^{2}\theta.
\]
Dakle, dobijamo
\[
m=\iiint\limits_{T}\rho(x,y,z)\,dx\,dy\,dz
=\int_{\pi}^{2\pi}\int_{0}^{\pi/2}\int_{3}^{4}
\bigl(3\rho^{2}\sin^{2}\theta\bigr)\cdot
\bigl(\rho^{2}\sin\theta\bigr)\,d\rho\,d\theta\,d\varphi.
\]
Granice su konstantne i promenljive međusobno nezavisne pa možemo razdvojiti na proizvode
\[
m=3\int_{\pi}^{2\pi}d\varphi
\int_{0}^{\pi/2}\sin^{3}\theta\,d\theta
\int_{3}^{4}\rho^{4}\,d\rho.
\]
Ovo su standardni integrali i dobijamo
\[
\int_{\pi}^{2\pi}d\varphi=\pi,\qquad
\int_{0}^{\pi/2}\sin^{3}\theta\,d\theta=\frac{2}{3},\qquad
\int_{3}^{4}\rho^{4}\,d\rho=\frac{4^{5}-3^{5}}{5}=\frac{781}{5}.
\]
Stoga je
\[
m=3\cdot\pi\cdot\frac{2}{3}\cdot\frac{781}{5}
=\frac{1562\pi}{5}.
\]
