# Thread: how do you draw paraboloid

1. ## Triple Integral

how do you draw paraboloid using z=1-x^2-y^2

sorry I haven't taken math for 2 years. I forget how to. I would really appreciated if you can explain me

this is the whole problem
use cylindrical coordinates. Evaluate ∫∫∫E (x^3 +xy^2)dV, where E is the solid in the first octant that lies beneath the paraboloid z= 1-X^2-y^2

I am trying to solve this problem but i feel like I won't know the bound until i know how to draw the paraboloid

2. To visualize the parabaloid, rewrite as $\displaystyle x^2+y^2=(\sqrt{1-z})^2$, and you can see that at a "height" z above the origin, we have a circle of radius $\displaystyle r=\sqrt{1-z}$. So we have a unit circle at $\displaystyle z=0$ coming to a point at $\displaystyle z=1$ to form a 3-D bell shape. To use just the first octant, we will quarter the bell much like you would quarter an apple, cutting it into four equal segments from the top. One of these segments represents the solid E.

Now convert to cylindrical coordinates:

$\displaystyle x=r\cos\theta$
$\displaystyle y=r\sin\theta$
$\displaystyle z=h$

$\displaystyle I=\int\int\int_E r^3(\cos\theta+\sin\theta)dE$, where $\displaystyle E=\{(x,y,z)\in\mathbb{R}^3|x>0,y>0,0<z<1-x^2-y^2\}$

To represent the region $\displaystyle E$ in cylindrical coordinates, start by visualizing $\displaystyle \theta$ ranging from $\displaystyle 0^\circ$ to $\displaystyle 90^\circ$ on our quartered bell, $\displaystyle r$ ranging from the "center" $\displaystyle r=0$ to the edge $\displaystyle r=1$. Finally, our inequality $\displaystyle z<1-x^2-y^2$ takes the form $\displaystyle h<1-r^2$, so a point at a distance $\displaystyle r$ from the z-axis can be at any height between $\displaystyle 0<h<1-r^2$. Therefore, $\displaystyle E=\{(r,\theta, h)\in\mathbb{R}^3|0<r<1,0<\theta<\frac{\pi}{2},0<h <1-r^2\}$. These are your bounds of integration.

Can you take it from there?