Prikaži rešenje
- Preseci krivih: iz
\[
\frac{x^{2}}{2}+y^{2}=1,\qquad x=y^{2}+\frac{1}{2}
\]
sledi \( y^{2}=\frac{1}{2}\Rightarrow y=\pm\frac{1}{\sqrt{2}} \) i \( x=1 \).
Tačke preseka: \( A\!\left(1,\frac{1}{\sqrt{2}}\right) \), \( B\!\left(1,-\frac{1}{\sqrt{2}}\right) \).
- Od dalje krive oduzimamo bližu:
\[
R(y)=\sqrt{2(1-y^{2})},\qquad r(y)=y^{2}+\frac{1}{2},
\]
\[
V=\pi\!\int_{-1/\sqrt{2}}^{1/\sqrt{2}}\!\big(R(y)^{2}-r(y)^{2}\big)\,dy
=2\pi\!\int_{0}^{1/\sqrt{2}}\!\left(\frac{7}{4}-3y^{2}-y^{4}\right)\!dy
\]
Rezultat: \( \frac{6\sqrt{2}}{5}\,\pi \)
4(a). Odrediti oblast definisanosti funkcije
\( f(x,y)=\ln(9-y^{2})+\frac{\sqrt{\,y-x+3\,}}{(x-1)^{2}+(y-2)^{2}}. \)
Prikaži rešenje
- \( 9-y^2 \gt 0 \Rightarrow -3 \lt y \lt 3 \).
- \( y-x+3\ge 0 \Rightarrow x\le y+3 \).
- \( (x-1)^2+(y-2)^2\ne 0 \Rightarrow (1,2) \) se isključuje.
Rezultat:
\( D=\{(x,y)\mid -3\lt y\lt 3,\ x\le y+3,\ (x,y)\ne(1,2)\} \)
4(b). Odrediti totalni diferencijal drugog reda funkcije \( z(x,y)=e^{y}\ln(x^{3}+2y) \) u \( M(1,2) \)
Prikaži rešenje
\[
d^{2}z=z_{xx}\,dx^{2}+2z_{xy}\,dx\,dy+z_{yy}\,dy^{2}.
\]
\[
z_x=e^{y}\frac{3x^{2}}{x^{3}+2y},\qquad
z_y=e^{y}\!\left[\ln(x^{3}+2y)+\frac{2}{x^{3}+2y}\right].
\]
\[
z_{xx}=e^{y}\,\frac{6x(x^{3}+2y)-9x^{3}}{(x^{3}+2y)^{2}},\qquad
z_{xy}=e^{y}\!\left[\frac{3x^{2}}{x^{3}+2y}-\frac{6x^{2}}{(x^{3}+2y)^{2}}\right],
\]
\[
z_{yy}=e^{y}\!\left[\ln(x^{3}+2y)+\frac{4}{x^{3}+2y}-\frac{4}{(x^{3}+2y)^{2}}\right].
\]
U tački \( M(1,2) \) je \( x^{3}+2y=5 \), pa
\[
z_{xx}(1,2)=\frac{21}{25}e^{2},\qquad
z_{xy}(1,2)=\frac{9}{25}e^{2},\qquad
z_{yy}(1,2)=\left(\ln 5+\frac{16}{25}\right)e^{2}.
\]
Rezultat: \( d^{2}z=\frac{21}{25}e^{2}\,dx^{2} + 2\cdot\frac{9}{25}e^{2}\,dx\,dy
+ \left(\ln 5+\frac{16}{25}\right)e^{2}\,dy^{2} \)
