Thread: Can someone check my work?

1. Can someone check my work?

Im studying for a final and I just need some reassurance.

1. Find the directional derivative of $\displaystyle f(x,y)=x^2+y^2+1 at (1,2) in the direction from (1,2) to (3,6)$

I found partial derivatives, found the direction vector, found the unit vector, My answer is $\displaystyle 16/sqrt (5)$

2.Find the area of the surface $\displaystyle z=16-x^2-y^2$ over the region $\displaystyle {(x,y)|x^2+y^2=16}$
I setup double integral using polar coordinates, r from 0 to 4, theta from 0 to 2pi.

2. The unit vector I obtain is $\displaystyle \frac{1}{\sqrt{5}}<1,2>$, the gradient is $\displaystyle <2x,2y>$ so the dot product is the function $\displaystyle D_uf(x,y)=2x\frac{1}{\sqrt{5}}+4y\frac{1}{\sqrt{5} }$, at the point you're given I'm obtaining $\displaystyle \frac{10}{\sqrt{5}}$

3. pshh..dur dur, I had an arithmetic brain fart. Yep thats what I have. lol

4. I'm obtaining a different answer for the surface area:

$\displaystyle S=\int_R\int\sqrt{1+[f_x(x,y)]^2+[f_y(x,y)]^2}dA$

$\displaystyle f_x(x,y)=-2x$
$\displaystyle f_y(x,y)=-2y$

$\displaystyle S=\int_R\int\sqrt{1+4x^2+4y^2}dA$

$\displaystyle =\int_{0}^{2\pi}\int_0^4\sqrt{1+4r^2}rdrd\theta$

$\displaystyle u=1+4r^2\rightarrow dr=8rdu$

$\displaystyle =\frac{1}{8}\int_0^{2\pi}\int_1^{65}\sqrt{u}dud\th eta$

$\displaystyle =\frac{1}{8}\frac{2}{3}\int_0^{2\pi}[u^{\frac{3}{2}}]_1^{65}d\theta$

$\displaystyle =\frac{1}{12}\int_0^{2\pi}(65\sqrt{65}-1)d\theta$

$\displaystyle =\frac{1}{12}(65\sqrt{65}\theta-\theta)]_0^\frac{2\pi}$

$\displaystyle \frac{65\sqrt{65}\pi-\pi}{6}$

5. Thanks, got it now. My brain is just on overload.