polar coordinates

Question

Suppose R₁,R₂ are real numbers satisfying 0 < R₁ < R₂. A function is defined in terms of polar coordinates by

f(r,θ) =√(R₁ + R₂)r−r² −R₁R₂.

 

(a) Show that the domain of this function is the region R = {(r,θ) : R₁ ≤ r ≤ R₂, 0 ≤ θ ≤ 2π}.

(b) Let T be the solid bounded by the (x,y)-plane and by the surface of (polar) equation z = f(r,θ).

      Calculate the volume of the solid.

 

Thank you

Answer 1

f(r,θ) =√(R₁ + R₂)r−r² −R₁R₂.

I assume that the whole (R₁ + R₂)r−r² −R₁R₂ is under the square root. (Actually, it should be that way!). If this is the case, then

(R₁ + R₂)r−r² −R₁R₂≥0

r2 − (R₁ + R₂)r+R₁R₂≤0

(r-R₁)(r-R₂)≤0.

From here it follows that R₁≤r≤R₂

As for θ, it doesn't appear implicitly in the expression for f(r, θ), so it can take its values on [0, 2π]. Thus the domain is {(r, θ): R₁≤r≤R₂, 0 ≤ θ ≤ 2π}.