Double Integrals in Polar Coordinates

iosa31
Messages
2
Reaction score
0

Homework Statement



Use polar coordinates to find the volume of the given solid.

Enclosed by the hyperboloid -x2 - y2 + z2 = 1 and the plane z = 2

Homework Equations



r2 = x2 + y2, x = rcosθ, y = rsinθ

∫∫f(x,y)dA = ∫∫f(rcosθ,rsinθ)rdrdθ

The Attempt at a Solution



-x2 - y2 + 4 = 1 → x2 + y2 = 3

0 ≤ r ≤ √3
0 ≤ θ ≤ 2∏

f(x,y) = √(1 + x2 + y2)
f(rcosθ,rsinθ) = √(1 + r2)

V = ∫∫r√(1 + r2)drdθ

u = 1 + r2
du = 2rdr

V = ∫∫1/2√ududθ
V = ∫1/3(u3/2)dθ
V = 1/3∫43/2 - 1dθ
V = 7/3∫dθ
V = (7/3)θ = 14∏/3
 
Physics news on Phys.org
Oops, I just realized the integral should be ∫∫(2 - √(1 + r^2))rdrdθ, not ∫∫r√(1 + r^2)drdθ. Sorry about that.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top