# Cal3 cyliderical spherical coords

• Eunix1992
So the volume is V = \int_{0}^{2 \pi} \int_{0}^{1} \int_{r}^{14 + r \cos \theta} \, dz \, dr \, d\theta .That's pretty ugly, but it's a standard calculus exercise to do the integration.In summary, the conversation discusses finding the volume between a cone and a plane, above a disk, using Cartesian or cylindrical coordinates. The approach is to express the cone as a function of z, and the plane as a function of r and theta. The integrand function for the height is determined and the volume is calculated using cylindrical coordinates. The range for z is between r

## Homework Statement

find the volume between the cone z=√(x^2+y^2), and the plane z=14+x, above the disk x^2+y^2≤1, for the exact number

r^2=x^2+y^2;

## The Attempt at a Solution

I found x=z, for x^2+y^2≤1, for solve r^2≤1, so r≤1, or r≥-1. for θ,from0 to 2pi, but I don't know what range for z, and what equation I shoud use for f(x,y), just √(x^2+y^2)?
thanks.

I think you are going to need to come down firmly on the side of working with either Cartesian or cylindrical coordinates, because this mixture of systems is creating unnecessary confusion.

You have an "apex-down" cone (the apex at ( 0, 0 ) ) opening "upward" in the positive z-direction being intersected by an oblique plane. You will find it easier to deal with this if you express the cone as z = f(r) (which will be the "floor" of your volume) and z = g( r, theta ) [what is 14 + x in polar coordinates?] (which is the "ceiling" of your volume).

You can now get an integrand function for the "height" of the volume as a function of r and theta . You will be integrating over 0 ≤ r ≤ 1 and you have the correct interval for theta. Be sure to use the infinitesimal volume elements for cylindrical coordinates. This will not be too hard to set up (it's a bit more work to actually integrate...).

dynamicsolo said:
I think you are going to need to come down firmly on the side of working with either Cartesian or cylindrical coordinates, because this mixture of systems is creating unnecessary confusion.

You have an "apex-down" cone (the apex at ( 0, 0 ) ) opening "upward" in the positive z-direction being intersected by an oblique plane. You will find it easier to deal with this if you express the cone as z = f(r) (which will be the "floor" of your volume) and z = g( r, theta ) [what is 14 + x in polar coordinates?] (which is the "ceiling" of your volume).

You can now get an integrand function for the "height" of the volume as a function of r and theta . You will be integrating over 0 ≤ r ≤ 1 and you have the correct interval for theta. Be sure to use the infinitesimal volume elements for cylindrical coordinates. This will not be too hard to set up (it's a bit more work to actually integrate...).

thanks.
but i still not understand the range for z, and that function i should use for f(x,y)which be intefrated. thanks.

I mean for z I used 14+x, and r; for r, I used 0 and 1; for theta, i use 0 and 2pi? if those are all correct, how can i get exact number? i should have x in final answer. Thanks

In polar coordinates (which are two of the three coordinates in the cylindrical system),
$z = 14 + r \cos \theta$ . So the height of the enclosed volume is $(14 + r \cos \theta ) - r$ , since the equation for the nappe of the cone above the xy-plane is just z = r .