Površina obrtnih tela — Zadaci

Izračunati površinu omotača tela nastalog obrtanjem krive \(y=\sin x\), za \(x\in[0,\pi/2]\), oko ose \(x\).

Za rotaciju oko ose \(x\): \(\displaystyle P=2\pi\int_a^b y\,\sqrt{1+(y’)^2}\,dx\).

Ovde \(y=\sin x\), \(y’=\cos x\): \[ P=2\pi\int_{0}^{\pi/2}\sin x\,\sqrt{1+\cos^2 x}\,dx. \] Neka \(u=\cos x\Rightarrow du=-\sin x\,dx\): \[ P=2\pi\int_{0}^{1}\sqrt{1+u^2}\,du =\pi\big(\sqrt{2}+\ln(1+\sqrt{2})\big). \]

Odgovor: \(P=\pi\big(\sqrt{2}+\ln(1+\sqrt{2})\big)\).

Naći površinu paraboloida dobijenog obrtanjem krive \(y=\sqrt{x}\), \(x\in[0,4]\), oko ose \(x\).

Za \(y=\sqrt{x}\) imamo \(y’=\frac{1}{2\sqrt{x}}\).

Formula za površinu pri obrtanju oko ose \(x\): \[ P=2\pi \int_0^4 y \,\sqrt{1+(y’)^2}\,dx. \]

Uvrstimo: \[ P = 2\pi \int_0^4 \sqrt{x}\,\sqrt{1+\frac{1}{4x}}\,dx = 2\pi \int_0^4 \sqrt{x}\,\sqrt{\frac{4x+1}{4x}}\,dx. \]

Posle sređivanja: \[ P = 2\pi \int_0^4 \frac{\sqrt{4x+1}}{2}\,dx = \pi \int_0^4 \sqrt{4x+1}\,dx. \]

Uvedimo smenu \(u=4x+1\), \(du=4\,dx\):

\[ P = \pi \int_{1}^{17} \frac{\sqrt{u}}{4}\,du = \frac{\pi}{4}\cdot \frac{2}{3}\,\left.u^{3/2}\right|_{1}^{17}. \]

\[ P = \frac{\pi}{6}\,\big(17^{3/2}-1\big) = \frac{\pi}{6}\,\big(17\sqrt{17}-1\big). \]

Odgovor: \(P=\frac{\pi}{6}\,(17\sqrt{17}-1)\).

Izračunati površinu tela nastalog rotacijom oko ose \(x\) krive određene uslovima \(x^{2}+y^{2}\le 18,\; x\le \frac{y^{2}}{3},\; y\ge 0.\)

Analiza regiona: Region je ograničen odozgo gornjim polukrugom \(y=\sqrt{18-x^2}\) (poluprečnik \(\sqrt{18}\)) i s desne strane parabolom \(x=\frac{y^2}{3}\). Kako je zadato \(y\ge 0\), posmatramo samo gornju polovinu. Tačke preseka dobijamo rešavanjem sistema:

\[ \frac{y^4}{9}+y^2=18 \;\Rightarrow\; y^2=9 \;\Rightarrow\; y=3,\;\; x=3. \]

1) Doprinos kružnice.
Kriva: \(y=\sqrt{18-x^2},\; x\in[-3\sqrt{2},\,3]\). Izvod: \(y’=-\frac{x}{\sqrt{18-x^2}}\). Tada \[ \sqrt{1+(y’)^2}=\frac{3\sqrt{2}}{\sqrt{18-x^2}}, \] pa je podintegralna funkcija konstanta \(3\sqrt{2}\): \[ P_{\text{kru}}=2\pi\!\int_{-3\sqrt{2}}^{3} 3\sqrt{2}\,dx =6\pi\sqrt{2}\,\left.x\right|_{-3\sqrt{2}}^{3} =\pi\,(36+18\sqrt{2}). \]

2) Doprinos parabole.
Parametrišemo po \(y\): \(x(y)=\frac{y^2}{3},\; y\in[0,3]\). Tada je \(\frac{dx}{dy}=\frac{2y}{3}\) i \[ P_{\text{par}}=2\pi\int_{0}^{3} y\,\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy =\frac{2\pi}{3}\int_{0}^{3} y\,\sqrt{9+4y^2}\,dy. \] Uz smenu \(u=9+4y^2,\; du=8y\,dy\) dobijamo \[ P_{\text{par}}=\frac{2\pi}{3}\cdot\frac{1}{8}\int \! u^{1/2}\,du =\frac{\pi}{12}\cdot\frac{2}{3}\,\left.u^{3/2}\right|_{y=0}^{y=3} =\frac{\pi}{18}\,\left.(9+4y^2)^{3/2}\right|_{0}^{3} =\frac{3\pi}{2}\,(5\sqrt{5}-1). \]

3) Zbir.
\[ P=P_{\text{kru}}+P_{\text{par}} =\frac{\pi}{2}\,\big(69+36\sqrt{2}+15\sqrt{5}\big). \]

Izračunati površinu tela nastalog rotacijom oko ose \(x\) dela parametarske krive \(x(t)=3\cos^{3}t,\; y(t)=3\sin^{3}t\) u drugom kvadrantu (npr. \(t\in[\pi/2,\pi]\)).

Za rotaciju oko ose \(x\) (parametar \(t\)): \[ P=2\pi\int y(t)\,\sqrt{\big(x'(t)\big)^2+\big(y'(t)\big)^2}\,dt. \] \(x’=-9\cos^{2}t\,\sin t,\;y’=9\sin^{2}t\,\cos t\Rightarrow \sqrt{x’^2+y’^2}=9\,|\sin t\,\cos t|\). U II kvadrantu \(\sin t\ge0,\ \cos t\le0\Rightarrow |\cos t|=-\cos t\).

\[ P=54\pi\!\int_{\pi/2}^{\pi}\sin^{4}t\,(-\cos t)\,dt =54\pi\!\int_{0}^{1}u^{4}\,du =54\pi\,\left.\frac{u^{5}}{5}\right|_{0}^{1} =\frac{54}{5}\,\pi. \]

Izračunati površinu tela nastalog rotacijom oko ose \(x\) zatvorene krive koju obrazuju parabole \(y=x^{2}\) i \(x=y^{2}\) u I kvadrantu (spajaju \((0,0)\) i \((1,1)\)).

1) Geometrija oblasti:
U intervalu \(x\in[0,1]\) gornja granica je \(y=\sqrt{x}\), a donja \(y=x^2\). Pri rotaciji oko ose \(x\) ukupna površina se dobija sabiranjem spoljašnjeg i unutrašnjeg omotača: \[ P = P_{\text{s}} + P_{\text{u}}. \]

2) Spoljašnji deo (\(y=\sqrt{x}\)):
Izvod je \[ y’ = \frac{1}{2\sqrt{x}}. \] Zbog toga \[ \sqrt{1+(y’)^2} = \sqrt{1+\frac{1}{4x}} = \frac{\sqrt{4x+1}}{2\sqrt{x}}. \] Pa je \[ y\sqrt{1+(y’)^2} = \sqrt{x}\cdot \frac{\sqrt{4x+1}}{2\sqrt{x}} = \frac{1}{2}\sqrt{4x+1}. \] Dakle \[ P_{\text{out}} = 2\pi \int_0^1 \frac{1}{2}\sqrt{4x+1}\,dx = \pi \int_0^1 \sqrt{4x+1}\,dx. \] Smena \(u=4x+1\), \(du=4dx\), daje: \[ P_{\text{out}} = \pi \int_{1}^{5} \frac{\sqrt{u}}{4}\,du = \frac{\pi}{4}\cdot \frac{2}{3}u^{3/2}\Big|_{1}^{5}. \] Konačno \[ P_{\text{s}} = \frac{\pi}{6}\,(5\sqrt{5}-1). \]

3) Unutrašnji deo (\(y=x^2\)):
Izvod je \[ y’ = 2x, \qquad \sqrt{1+(y’)^2}=\sqrt{1+4x^2}. \] Formula daje \[ P_{\text{in}} = 2\pi \int_0^1 x^2\sqrt{1+4x^2}\,dx. \] Smena \(u=2x\), \(dx=\frac{du}{2}\), \(x=\frac{u}{2}\): \[ \int_0^1 x^2\sqrt{1+4x^2}\,dx = \int_0^2 \left(\frac{u^2}{4}\right)\sqrt{1+u^2}\cdot \frac{du}{2} = \frac{1}{8}\int_0^2 u^2\sqrt{1+u^2}\,du. \] Dalje, neka \(t=1+u^2\), \(dt=2u\,du\), pa: \[ \int u^2\sqrt{1+u^2}\,du = \frac{1}{2}\int (t-1)\sqrt{t}\,dt = \frac{1}{2}\int (t^{3/2}-t^{1/2})\,dt. \] \[ \frac{1}{2}\left(\frac{2}{5}t^{5/2} – \frac{2}{3}t^{3/2}\right) = \frac{1}{5}t^{5/2} – \frac{1}{3}t^{3/2}. \] \[ \int_0^2 u^2\sqrt{1+u^2}\,du = \left(\frac{1}{5}t^{5/2}-\frac{1}{3}t^{3/2}\right)\Big|_{1}^{5}. \] Nakon sređivanja i vraćanja konstanti: \[ \int_0^1 x^2\sqrt{1+4x^2}\,dx = \frac{1}{8}\Big(\frac{18\sqrt{5}}{5}-\frac{1}{3}\Big). \] Tako dobijamo \[ P_{\text{u}} = 2\pi\cdot \frac{1}{8}\Big(\frac{18\sqrt{5}}{5}-\frac{1}{3}\Big) = \pi\left(\frac{9}{16}\sqrt{5}-\frac{1}{32}\ln(2+\sqrt{5})\right). \]

4) Rezultat:
\[ P = P_{\text{s}}+P_{\text{u}} = \frac{\pi}{6}(5\sqrt5-1) + \pi\left(\frac{9}{16}\sqrt5-\frac{1}{32}\ln(2+\sqrt5)\right). \] Posle sređivanja: \[ P = \pi\left(\frac{67}{48}\sqrt{5}-\frac{1}{6}-\frac{1}{32}\ln(2+\sqrt{5})\right). \]