Find the volume of the solid that lies within both the cylinder x²+ y²= 1 and the sphere x²+ y²+ z²= 4

Solution:

Given,

Equation of cylinder is x²+ y²= 1

Equation of sphere is x²+ y²+ z²= 4

In cartesian coordinates,

the region is given by -1 ≤ x ≤ 1, - √1 - x² ≤ y ≤ √1 - x²

and - √4 - x² - y² ≤ y² ≤ √4 - x² - y²

Converting to cylindrical coordinates, using

∫x(r,θ,𝛾) = r cosθ

∫y(r,θ,𝛾) = r sinθ

∫z(r,θ,𝛾) = 𝛾

We get a Jacobian determinant of r,

The region is converted into cylindrical coordinates by 0 ≤ θ ≤ 2𝜋, θ ≤ r ≤ 1 and - √4 - r² ≤ z ≤ √4 - r².

Using integration to find the volume,

\(\\Volume=\int_{\Theta =0}^{\Theta =2\pi }\int_{r=0}^{r=1}\int_{z=-\sqrt{4}-r2}^{z=\sqrt{4}-r2}r dz dr d\Theta \\ \\Volume=\int_{\Theta =0}^{\Theta =2\pi }\int_{r=0}^{r=1}2r \sqrt{(4-r^{2})}drd\Theta \\ \\Volume=\int_{\Theta =0}^{\Theta =2\pi }-2(4-r^{2})^{3/2}/3d\Theta \\ \\Volume=\int_{\Theta =0}^{\Theta =2\pi }\frac{16}{3}-2\sqrt{3}d\Theta\)

= [θ(16/3 - 2√3)] in the limit θ = 0 to 2𝜋

= [2θ(8 - 3√3) / 3] in the limit θ = 0 to 2𝜋

= [2(2𝜋) (8 - 3√3)/3 - 0]

= 4𝜋 / 3 (8 - 3√3)

Therefore, the volume is 4𝜋 / 3 (8 - 3√3) units.