I am struggling with this:
Find the surface area of the patch of the graph $z = -x^5-y^2+e^{xy-1}$ for $1\leqq x \leqq 1.1$ and $1\leqq y \leqq 1.2$
I am unsure of how to go about solving this.
I have already tried solving it by parametrizing the equation in terms of $r$ and $\theta$, then taking double integral of the cross products of $F_r$ and $F_\theta$. However, I am unsure of how to find the bounds of $r$ after.
I'm looking for an approximate result.
Also, I tried solving while keeping it in $xy$ coordinates, however I get a problem when solving for $dS$. The $dS$ I get seems a bit too complicated than what seems to be intended, and I feel i'm missing an important step somewhere. I'm guessing perhaps this has to do with whether I'm looking for an approximate or exact result.
Considering that the size of the patch, a 0.1 by 0.2 rectangle, is rather small, an analytic approximate for the surface integral over this small space can be derived
Give that the upper limits $x=1.1$ and $y=1.2$ are both close to 1, the function $z$ can be expanded around $x=y=1$. To the first order in $u=x-1$ and $v=y-1$, $z$ can be approximated as,
$$z(u,v) = -(1+u)^5-(1+v )^2 +e^{(1+u)(1+v)-1}\approx -1 - 4u-v$$
Then, $z_u’= -4$ and $z_v’= -1$, and the surface integral is,
$$S= \int_0^{0.2}\int_0^{0.1} \sqrt{1+(z_u’)^2+(z_v’)^2}dudv$$ $$=\int_0^{0.2}\int_0^{0.1}\sqrt{18}dudv=0.06\sqrt{2}$$