# Thread: Evaluate using spherical coordinates.

1. ## Evaluate using spherical coordinates.

Evaluate using spherical coordinates: $\displaystyle \int \int \int dx dy dz$
$\displaystyle T: 0 \le x \le 1 \; \; ; \; \; 0 \le y \le \sqrt{1-x^2} \; \; ;\; \; \sqrt{x^2 + y^2} \le z \le \sqrt{2 - (x^2 + y^2)}$

I can represent this in cartesian coordinates like this:

$\displaystyle \int_0^1 \int_0^{\sqrt{1-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{2-(x^2+y^2)}}dzdydx$

and from here convert to spherical coordinates. I'm not sure how to completely convert this to spherical coordinates.

The only thing I could see was that on the xy-plane, part of the area is a quarter of a circle in the first quadrant with radius 1 (since y goes from 0 to $\displaystyle \sqrt{1-x^2}$ and x goes from 0 to 1), so $\displaystyle \theta$ would go from 0 to $\displaystyle \frac{\pi}{2}$.

2. ## Re: Evaluate using spherical coordinates.

drawing the area will help you how I draw it first I begin with the x-axis x changes from 0 to 1 as the drawing

then we will draw y , y changes from 0 to sqrt{1-x^2} . sub x=0 and x=1 which will give us quarter of circle in xy plane in the positve for xy

now z the hardest one first let see what is sqrt{x^2+y^2} = z if x=0 , y=1 , z=1 and if y=0,x=1,z=1
if z=1 this will give us a circle in xy plane with radius 1 but we interested just in xy positive

now we will see z= sqrt{2-x^2-y^2 } , if x=0,y=0, then z=sqrt{2}
it is sphere but positive z half of the sphere but since x,y are positive too so it is the quarter of the sphere
z=sqrt{2-x^2-y^2} will intersect with z=sqrt{x^2+y^2} at z=1 the circle in xy plane

the area is the area with the purple color
now can you determine the boundaries ?

3. ## Re: Evaluate using spherical coordinates.

$\displaystyle z= \sqrt{x^2+ y^2}$ is the part of $\displaystyle z^2= x^2+ y^2$ above the xy-plane. As an aid to seeing what that is, note that when y= 0, it gives $\displaystyle z^2= x^2$ or $\displaystyle z= \pm x$ and similarly for y. That is, this is a cone making an angle 0f $\displaystyle \pi/4$ with the z-axis. The other boundary, $\displaystyle z= \sqrt{2- x^2- y^2}$ is the part of $\displaystyle z^2= 2- x^2- y^2$ or $\displaystyle x^2+ y^2+ z^2= 2$ above the xy-plane. That is, of course, a sphere of radius $\displaystyle \sqrt{2}$. Yes, $\displaystyle \theta$ goes from 0 to $\displaystyle \pi/2$. $\displaystyle \rho$ goes from 0 to $\displaystyle \sqrt{2}$ and $\displaystyle \phi$ from 0 to $\displaystyle \pi/4$.