Answer:
[tex]4/3(2\sqrt{2} -1)*\pi[/tex] this is the answer
Step-by-step explanation:
[tex]1<= x^2+y^2 <= 2 by the cons z= +- \sqrt{(x^2 + y^2)}[/tex]
Let suppose
x = r∙cos(θ)
y = r∙cos(θ)
z = z
The differential volume element changes to
dV = dxdydz = r drdθdz
The limits of integration in cylindrical coordinates are:
The limits of integration in cylindrical coordinates are:
(i)
[tex]1 \leq x^2 + y^2\leq 2[/tex]
[tex]1 \leq r^2 \leq 2[/tex]
since r is always positive
[tex]1 \leq r \leq \sqrt{2}[/tex]
(ii)
[tex]- \sqrt{(x^2+ y^2)} \leq z \leq +\sqrt{(x^2+ y^2)}[/tex]
[tex]- r \leq z \leq r[/tex]
(iii)
[tex]0 \leq Theta \leq 2∙π[/tex]
we have no restrictions in radial direction.
[tex]V = \int\limits^a_b { dV} \,[/tex]
Remaining derivation has been explained in the atatchment where we get the volume of the cylinder
