Find the volume of the given solid:

[Bula]

Homework Statement

The solid is above the paraboloid z=x^2+y^2 and below the half-cone z=√(x^2+y^2)

The Attempt at a Solution

First off, I would like to apologize if this question has already been posted. I'm new to the forum.

I drew the two solids. According to the picture I drew, it seems as if I'm looking for the volume of the cone?

Setting the two equations equal to one another, I find that p=1.

The equation I've set up (pretty sure it's completely wrong).
int(0,2pi)int(0,pi/4)int(0,1)$\rho^2sin(\phi)$$d\rho$$d\phi$$d\theta.$Yeah, so it's pretty clear I'm clueless of what I'm doing. Any help would be appreciated.
Thank you.

 

[DryRun]

Where's the drawing? You need to use cylindrical coordinates. The volume that you're trying to find is sandwiched between the cone and the paraboloid.

The answer is... $\frac{\pi}{6}$?

 

[tiny-tim]

Welcome to PF!

Hi Bula! Welcome to PF!

I drew the two solids. According to the picture I drew, it seems as if I'm looking for the volume of the cone?

Nooo … draw it in 2D …

it's the area between the line z = r and the parabola z = r², rotated about the z axis.

Try again! ​

 

[Bula]

Where's the drawing? You need to use cylindrical coordinates. The volume that you're trying to find is sandwiched between the cone and the paraboloid.

The answer is... $\frac{\pi}{6}$?

Hi Bula! Welcome to PF! Nooo … draw it in 2D …

it's the area between the line z = r and the parabola z = r², rotated about the z axis.

Try again! ​

Thank you for the warm welcome and replies.

Unfortunately, I'm still unsure of how to solve this problem.
http://www.google.com/imgres?hl=en&biw=1440&bih=785&gbv=2&tbm=isch&tbnid=TM-t-hB2wyczZM:&imgrefurl=http://docs.racket-lang.org/plot/intro.html&docid=S_HCHWyaV06zOM&imgurl=http://docs.racket-lang.org/plot/pict_5.png&w=400&h=400&ei=3TGLT52nJcq30AGyweXWCQ&zoom=1&iact=rc&dur=406&sig=100761004641933490948&page=2&tbnh=139&tbnw=139&start=29&ndsp=35&ved=1t:429,r:2,s:29,i:170&tx=87&ty=53
Ignoring the x=y^2 graph, the area we're trying to find is that in the first quadrant, right?

If it's not to much to ask, could you provide the integral? It's easier for me to work backwards.

EDIT: Would the integral be;
int(0,2pi)int(0,1)int(0,r) (r-r^2)dzrdrd$\theta$?

 

[DryRun]

You need to do a complete 2pi revolution about the z-axis to get the volume. You should definitely draw the surfaces so you'll know exactly what you're being asked to do. Start with an xy coordinate trace, zy and then zx. Then draw the whole thing in 3D. Visualize it so you'll understand the limits, which are the heart of the problem here. Project the object onto the xy plane, so you'll get the points of intersection (the equation of a circle) of the cone and the paraboloid.

 

[tiny-tim]

Hi Bula!

(have an integral: ∫ and a theta: θ and try using the X² button just above the Reply box )

EDIT: Would the integral be;
int(0,2pi)int(0,1)int(0,r) (r-r^2)dzrdrd$\theta$?

You're confusing the limits with the integrand.

The integrand (except for the r in drdzdθ) is only the density (which in this case is just 1).

The r and r² (or z and √z) should appear somewhere in the limits.