Find the volume between x^2+y^2=z^2 and x^2+y^2=z. Just want some clarifications. Ans. (pi/6)?
I answered this by converting to polar coordinates. I set the equation as:
[0, 2pi], [0,1], [r^2, r] r dz dr d(theta). I got the correct answer.
However, I am confused about the boundaries of "z."
At first, I used [r, r^2] and got negative.
What is the basis in what will become the lower or upper boundary in this?
Sorry, I'm just new to this
- Steve4PhysicsLv 74 months agoFavorite Answer
That’s the danger of not ‘visualising’ what you are trying to find. [r, r^2] is back-to-front. I’ll walk you through using my preferred method. Warning long answer. You should see what’s going on.
In cylindrical polar coordinates the equations are:
r² = z² (equation 1, giving a surface S1)
r² = z (equation 2, giving a surface S2)
These 2 surfaces intersect when z² = z which gives z = 0 and z = 1.
S1 is an inverted cone (tip at origin); the radius of each circular cross section is z.
S2 is an inverted cone-like surface (tip at origin); the radius of each circular cross section is √z. So the sides of the S2 ‘cone’ are curved.
We want the volume between surfaces S1 and S2. We need to work out which one is on the outside.
In the interval 0≤z≤1 note that z≤√z. For example when z=1/2, √z = √(1/2) = 0.707 which is bigger than z. So S2 is *outside* and S1 is *inside*. We want the volume between S1 and S2.
For a given value of z between z=0 and z=1, a cross section of the volume is an annulus with outer radius ro = √z and inner radius ri = z. So the area of the annulus = πro² - πri² = πz – πz² = π(z – z²). Volume of annulus is π(z – z²)dz.
Volume is the sum of all these annuli:
V = ∫π(z – z²)dz evaluated from z=0 to z=1
= π[z²/2 – z³/3] evaluated from z=0 to z=1
= π[1²/2 – 1³/3]
- RealProLv 74 months ago
If the cone is above the parabola then the equation for the cone is the upper bound. If you want to find the area between y=x and y=x^2 you will integrate x - x^2 between 0 and 1 and not x^2 - x, right?