Analys II · 2026-08-24
$f(x,y)=e^{x^2+y^2}$.
Beräkna dubbelintegralen $\iint_D f(x,y)\,dx\,dy$, där $D=\{(x,y):y\ge\lvert x\rvert,\ x^2+y^2\le1\}$.
Låt $g(r,t)=f(2r\cos t,r\sin t)$. Beräkna $\frac{\partial g}{\partial r}$ och $\frac{\partial g}{\partial t}$ med kedjeregeln i två variabler. Bestäm därefter tangentplanet till ytan $s=g(r,t)$ i punkten $(0,0,1)$.
(i) Området
$$D=\{(r,\theta):r\le1,\ \pi/4\le\theta\le3\pi/4\}.$$
Vi får
$$\iint_Df(x,y)\,dx\,dy
=\int_{\pi/4}^{3\pi/4}\left(\int_0^1e^{r^2}r\,dr\right)d\theta
=\frac\pi4(e-1).$$
(ii) Vi har
$$\frac{\partial f}{\partial x}=2xe^{x^2+y^2},\qquad
\frac{\partial f}{\partial y}=2ye^{x^2+y^2},$$
och
$$\frac{\partial x}{\partial r}=2\cos t,\quad
\frac{\partial y}{\partial r}=\sin t,\quad
\frac{\partial x}{\partial t}=-2r\sin t,\quad
\frac{\partial y}{\partial t}=r\cos t.$$
Därför
$$\frac{\partial g}{\partial r}
=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}
+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}
=2r(4\cos^2t+\sin^2t)e^{r^2(4\cos^2t+\sin^2t)},$$
$$\frac{\partial g}{\partial t}
=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}
+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}
=-6r^2\sin t\cos t\,e^{r^2(4\cos^2t+\sin^2t)}.$$
Tangentplanet till $s=g(r,t)$ i $(0,0,1)$ är
$$s-1=\frac{\partial g}{\partial r}(0,0)\,r+
\frac{\partial g}{\partial t}(0,0)\,t,$$
dvs. $s=1$.