# Thread: Domain of 3-dimensional Surfaces

1. ## Domain of 3-dimensional Surfaces

Hi,

I was wondering how one would go about finding the domain of a 3-dimensional surface. For example, if the equation was f(x,y) = (sqrt(x + 2y + 1)) * (sqrt(x - 2y + 1)) * (ln(1-sqrt(tan(x^2 + 2x - 4y^2)))) what would the domain be? Thanks!

Hi,

I was wondering how one would go about finding the domain of a 3-dimensional surface. For example, if the equation was f(x,y) = (sqrt(x + 2y + 1)) * (sqrt(x - 2y + 1)) * (ln(1-sqrt(tan(x^2 + 2x - 4y^2)))) what would the domain be? Thanks!
there will be three coditions

$x+2y+1>0$

$x-2y+1>0$

$0< tan(x^2 + 2x - 4y^2)<1$

Hi,

I was wondering how one would go about finding the domain of a 3-dimensional surface. For example, if the equation was f(x,y) = (sqrt(x + 2y + 1)) * (sqrt(x - 2y + 1)) * (ln(1-sqrt(tan(x^2 + 2x - 4y^2)))) what would the domain be? Thanks!
$f(x,y)=\left[\sqrt{x+2y+1}\right]\left[\sqrt{x-2y+1}\right]\left[\ln(1-\sqrt{tan(x^2+2x-4y^2)}\right]$

find the domain for the first and the second and third each one alone then find the intersect of the result you have

you should know that the square root of the negative numbers is undefined and ln the natural log is the same and for the tan(s) ''s is arbitrary function'' cos(s) should not equal zero

$x+2y+1 \geq 0$ the domain of the first one for the second

$x-2y+1 \geq 0$ for the third

$1-\sqrt{tan(x^2+2x-4y^2)}> 0$ so

$1>\sqrt{tan(x^2+2x-4y^2)}$ and

$tan(x^2+2x-4y^2) \geq 0$ tan is positive in the first and the third quarter so

$0\leq tan(t)<1$

if $0\leq t< \frac{\pi}{4}$

this in the first quarter in the third

$\pi \leq t <\frac{\pi}{4}+\pi$ in general

$n\pi \leq t < \frac{\pi}{4}+n\pi$ so

$n\pi \leq x^2+2x-4y^2 < \frac{\pi}{4} +n\pi$ n integer

$(x+2y+1)(x-2y+1)\geq 0$ since the both of them is positive

$x^2-4y^2+2x+1 \geq 0$

$x^2-4y^2+2x \geq -1$ forget it

the domain is

$\left(0 \leq x^2+2x-4y^2 < \frac{\pi}{4}\right)\bigcup \left(\pi \leq x^2+2x-4y^2 < \frac{\pi}{4} + \pi\right) \bigcup$ $\left( 2\pi \leq x^2+2x-4y^2 < \frac{\pi}{4}+2\pi\right)\bigcup .....$

that is the way you can find the domain