Projekat iz Matematike 2 – Tema 18

Rešenje:
Kriva L je presečna linija dve površine. Parametrizacija se dobija uvođenjem pogodnih koordinata (npr. polarnih) ili direktno iz uslova.
\( x^2 + y^2 + 2 = 8 – \sqrt{x^2 + y^2} \)
\( r^2 = x^2 + y^2 r^2 + 2 = 8 – r r^2 + r – 6 = 0 r=2 \)

Računa se element dužine luka ds.
\( x = 2\cos t y = 2\sin t z = 6t \in (0,2\pi) ds = 2dt \)

Množi se sa gustinom ρ(x,y,z).
\( \rho = 4\cos^2 t – 2 \)
Postavlja se integral mase po parametru.Računa se integral.
\( M = \int_0^{2\pi} (4\cos^2 t – 2)2dt = \int_0^{2\pi} (8\cos^2 t – 4)dt \)
\( \cos^2 t = \frac{1+\cos 2t}{2}\)
\( M = \int_0^{2\pi} (4 + 4\cos 2t – 4)dt = \int_0^{2\pi} 4\cos 2tdt = 0 \)

U ovom zadatku rezultat je M = 0, jer se gustina duž krive izražava tako da se članovi u integralu međusobno ponište.

\( M = 0 \)

Postupak

Krajnje rešenje

\[\int_{0}^{4}\left(\int_{\sqrt{4-x}}^{2} f(x,y)\,dy\right)dx.\]Rešenje: Oblast integracije:\[D=(x,y)\mid 0\le x\le4,\ \sqrt{4-x}\le y\le2. \]Kvadriranjem nejednakosti (pošto su obe strane nenegativne) dobijamo:\[\sqrt{4-x}\le y \Longleftrightarrow 4-x\le y^2 \Longleftrightarrow x\ge 4-y^2.\]Opseg za y: iz granica sledi \(0\le y\le2\). Za svaki \(y\) u tom intervalu, \(x\) ide od \(4-y^2\) do \(4\). Dakle promenjeni red integracije je:\[\boxed{\displaystyle\int_{0}^{2}\left(\int_{4-y^{2}}^{4} f(x,y)\,dx\right)dy.}\]

\[\boxed{\displaystyle\int_{0}^{2}\left(\int_{4-y^{2}}^{4} f(x,y)\,dx\right)dy.}\]

Zadatak:
$$I = \iint\limits_D (1+4x+y)(7x+4y)\, dxdy,$$gde je oblast D ograničena pravama:$$y=-4x, \quad y=-4x+1, \quad x=2y-3, \quad x=2y-1.$$
Uvodimo novu promenljivu:$$u = 1+4x+y, \quad v = 7x+4y.$$Tada:$$f(x,y) = (1+4x+y)(7x+4y) = uv.$$
$$\frac{\partial(u,v)}{\partial(x,y)} =\begin{vmatrix}4 & 1 \\7 & 4\end{vmatrix} = 16-7=9,$$pa je:$$dxdy = \tfrac{1}{9}\, dudv.$$
Za y=-4x: u=1, Za y=-4x+1: u=2, Za x=2y-3: v=2u+1, Za x=2y-1: v=2u-1.
Dakle:$$1 \leq u \leq 2, \quad 2u-1 \leq v \leq 2u+1.$$
$$I = \frac{1}{9}\int_1^2 \int_{2u-1}^{2u+1} uv \, dv\,du.$$
Računamo unutrašnji integral:$$\int_{2u-1}^{2u+1} uv \, dv= \frac{u}{2}\left((2u+1)^2-(2u-1)^2\right)= 4u^2.$$ Zato:$$I = \frac{1}{9}\int_1^2 4u^2\,du= \frac{4}{9}\cdot \frac{7}{3}= \frac{28}{27}.$$

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\[\boxed{I = \tfrac{28}{27}}\]

$$\iiint\limits_G dz\,dy\,dx,$$gde je oblast$$G = (x,y,z) \mid z \leq \sqrt{9x^2+y^2}-3, \; z \geq 9-9x^2-y^2 .$$ $$z \leq \sqrt{9x^2+y^2}-3 \quad \Rightarrow \quad z+3 \leq \sqrt{9x^2+y^2},$$pa važi:$$(z+3)^2 \leq 9x^2+y^2.$$Takođe:$$z \geq 9-9x^2-y^2.$$ $$V = \iiint\limits_G dz\,dy\,dx= \iint\limits_D \left(\sqrt{9x^2+y^2}-3 - (9-9x^2-y^2)\right) dy\,dx,$$odnosno$$V = \iint\limits_D \left(\sqrt{9x^2+y^2}-12+9x^2+y^2\right)\,dy\,dx.$$ $$V = \int_0^{2\pi}\int_0^3 (3r-12+9r^2)\,r\,dr\,d\theta.$$ $$V = \int_0^{2\pi} d\theta \int_0^3 (3r^2+9r^3-12r)\,dr= 2\pi \int_0^3 (9r^3+3r^2-12r)\,dr.$$Sada:$$\int_0^3 9r^3\,dr = \left.\frac{9r^4}{4}\right|_0^3 = \frac{9\cdot 81}{4} = \frac{729}{4},$$$$\int_0^3 3r^2\,dr = \left.r^3\right|_0^3 = 27,$$$$\int_0^3 -12r\,dr = \left.-6r^2\right|_0^3 = -54.$$Ukupno:$$\int_0^3 (9r^3+3r^2-12r)\,dr = \frac{729}{4}+27-54 = \frac{729}{4}-27 = \frac{621}{4}.$$Zato:$$V = 2\pi \cdot \frac{621}{4} = \frac{621\pi}{2}.$$

---\[\boxed{V = \tfrac{621\pi}{2}}\]

Površina je paraboloid: \[z = 1 + x^2 + y^2.\]Za računanje površinskog integrala koristimo formulu: \[\iint_{S} f(x,y,z)\, dS = \iint_{D} f(x,y,z(x,y)) \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2}\, dxdy.\]Imamo: \[z_x = 2x, \quad z_y = 2y, \quad \sqrt{1+z_x^2+z_y^2} = \sqrt{1+4x^2+4y^2}.\]Takođe: \[f(x,y,z) = \sqrt{4z-3} = \sqrt{4(1+x^2+y^2)-3} = \sqrt{4x^2+4y^2+1}.\]Dakle: \[f(x,y,z)\,\sqrt{1+z_x^2+z_y^2} = 4x^2+4y^2+1.\]Integral prelazi u: \[I = \iint_{D} (4x^2 + 4y^2 + 1)\, dxdy.\]Ravni: \[z = 1 + y, \quad z = 1 + 2y.\]Sečenjem sa paraboloidom dobijamo kružnice: \[x^2+y^2 = y, \quad x^2+y^2 = 2y.\]U polarnim koordinatama: \[r = \sin\theta, \quad r = 2\sin\theta, \quad 0 \leq \theta \leq \pi.\]Dakle \(D\) je prsten između ove dve kružnice.\[I = \int_0^\pi \int_{\sin\theta}^{2\sin\theta} (4r^2+1)\,r\,dr\,d\theta.\]Unutrašnji integral: \[\int (4r^2+1)r\,dr = r^4 + \tfrac{1}{2}r^2.\]Na granicama: \[\Big(r^4 + \tfrac{1}{2}r^2\Big)_{\sin\theta}^{2\sin\theta} = 15\sin^4\theta + \tfrac{3}{2}\sin^2\theta.\]Računanje:\[I = \int_0^\pi \left(15\sin^4\theta + \tfrac{3}{2}\sin^2\theta\right)\,d\theta.\]Poznati integrali: \[\int_0^\pi \sin^2\theta\,d\theta = \tfrac{\pi}{2}, \quad \int_0^\pi \sin^4\theta\,d\theta = \tfrac{3\pi}{8}.\]Zamenom dobijamo: \[I = 15 \cdot \tfrac{3\pi}{8} + \tfrac{3}{2}\cdot \tfrac{\pi}{2} = \tfrac{45\pi}{8} + \tfrac{6\pi}{8} = \tfrac{51\pi}{8}.\]

\[I = \frac{51\pi}{8}\]