Spherical Coordinates Question

[meeatingrice]

Hi, I am having trouble starting off this question. Could someone help me start off? Thanks in advance...

Use spherical coordinates to evaluate
∫∫∫н (x² + y²) dV,
where H is the hemispherical region that lies above the x-y plane and below the sphere x² + y² + z² = 1.

 

[HallsofIvy]

In spherical coordinates, $x=\rho cos(\theta)sin(\phi)$, $y= \rho sin(\theta)sin(\theta)$, z= $\rho cos(\phi)$ so
$x^2+ y^2= \rho^2 sin^2(\theta)$, $dV= \rho^2 sin(\phi)d\rho d\theta d\phi$, and the limits of integration are $\rho$ from 0 to 1, $\theta$ from 0 to $2\pi$, and $\phi$ from 0 to $\pi/2$.

 

[meeatingrice]

In spherical coordinates, $x=\rho cos(\theta)sin(\phi)$, $y= \rho sin(\theta)sin(\theta)$, z= $\rho cos(\phi)$ so
$x^2+ y^2= \rho^2 sin^2(\theta)$, $dV= \rho^2 sin(\phi)d\rho d\theta d\phi$, and the limits of integration are $\rho$ from 0 to 1, $\theta$ from 0 to $2\pi$, and $\phi$ from 0 to $\pi/2$.

Thanks HallsofIvy for the reply. So is this basically is this just finding a volume to the hemisphere that's below the x-y plane? Sorry with the questions, I"m new to this...

 

[HallsofIvy]

Thanks HallsofIvy for the reply. So is this basically is this just finding a volume to the hemisphere that's below the x-y plane? Sorry with the questions, I"m new to this...

No, it's not. For one thing, you said it was the portion of the sphere above the x-y plane, not below. More importantly, it's not finding the volume because that would be just $\int \int \int dV$, not $\int \int \int (x^2+ y^2) dV$. You are integrating the function x²+ y² over the unit hemisphere above the xy-plane.

 

[BlkDaemon]

Clarification, from a casual observer

Is sine supposed to be squared in dV? If not, why?

Also, what's the difference between "evaluating a hemisphere" as stated in the original problem and "finding volume"? When I took VectorCalc, this kind of stuff used to stump me all the time.

And lastly, how do you know what the limits are for each variable, e.g. 0 to 1, 0 to 2pi, etc?

 

[HallsofIvy]

the sin² was my error. I have edited it.

The problem did not say "evaluate a hemisphere" (I don't know what that could mean). It said "evaluate the integral" on the hemisphere.
Surely you know what it means to "evaluate an integral".

 

[SLChewy]

since x=psin(thie)cos(theta) and y= psin(thie)sin(theta) squaring both quantanties would give you p^2sin^2(thie)cos^2(theta) + psin^2sin(thie)sin^2(theta) and using an identity give yous p^2sin^2(thie) + p^2sin^2(thie) since the theta's would equal one. Then adding these together this would give you 2p^2sin^2(thie) wouldn't it. then multiplying by p^2sin(thie) give you the thing you integrate first by rho, then theta, and finaly thie right. This is problem18 on p887 of James Stewart Calculus Concepts and Contexts 3