# Linear approximation and Multiple integral questions

1. Jul 12, 2012

### fletch-j

I am aware that for a function of two variables $f(x,y)$ a linear approximation of a point $f(x,y)$ close to $f(x_0,y_0)$ can be approximated by the tangent plane approximation $f(x_0+\Delta x,y_0+\Delta y)\approx f(x_0,y_0)+f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y$ where $\Delta x=x-x_0$ and $\Delta y=y-y_0$

So is the process the same if you have a function $f(x,y)$ where $x$ and $y$ are themselves functions of more variables?
For example approximating $f(x(s,t),y(s,t))$ near $f(x(s_0,t_0),y(s_0,t_0))$

Does it scale to an arbitrary number of nested functions?

Would it be something like:

$f(x(s,t),y(s,t)) \approx f(x(s_0,t_0)+x_s(s_0,t_0)\Delta s+x_t(s_0,t_0)\Delta t , y(s_0,t_0)+y_s(s_0,t_0)\Delta s+y_t(s_0,t_0)\Delta t)$

or

$f(x(s,t),y(s,t)) \approx f(x(s_0,t_0),y(s_0,t_0))+f_s(x(s_0,t_0),y(s_0,t_0))\Delta s + f_t(x(s_0,t_0),y(s_0,t_0))\Delta t$
??

Also I have this problem:

1. The problem statement, all variables and given/known data
Use a triple integral to find the volume of a solid cut from $x=y^2$ by the planes $z=0$ and $x+z=1$

I just need help setting it out, I can do the calculation myself.

3. The attempt at a solution

I had it set out as follows but I am unsure if it is correct.

$\int^1_{-1} \int^1_0 \int^{1-z}_{y^2} dx dz dy$

Can someone please let me know if that is incorrect and if so, what I have done wrong.

Also I was wondering, (how) could this be written as a double integral?

Last edited: Jul 12, 2012
2. Jul 12, 2012

### LCKurtz

You have chosen the worst possible order of integration for this problem, which has caused you to get the middle limits incorrect. The difficulty is that if you integrate in the x direction first, you need to know the shadow of the solid in the yz plane. To get that you need to eliminate x between the cylindrical parabola and the slanted plane. The upper limit on the dz integral is not constant since it depends on y.

The easiest order would be dz dx dy. Once you work out the inner integral, what is left is how you would set the problem up as a double integral.

3. Jul 13, 2012

### fletch-j

Ah crap, that makes perfect sense. Thanks for being blunt.