Question

Use the triple integrals and spherical coordinates to find the volume of the solid that is bounded by the graphs of the given equations. x^2+y^2=4, y=x, y=sqrt(3)x, z=0, in first octant.

Answer 1

Solution

Let us assume the upper limit of z be 2

So limits of would be

equation of cylinder in spherical coordinates would be,

sin2 4>p = 2 csc)

So limits of p would be 0 2 csc

and limits of would be <

$\text{Therefore, the volume will be, }$

2 csc p sin()dpdodo V =

312 csc sin()dod 3 0 프 프

$\text{V}=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left [ \frac{8\csc^3 \phi }{3} \right ]\sin(\phi) d\phi d\theta$

2 8 csc 2 dodo 3 V =

$\text{V}=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\left [ -\dfrac{8}{3\tan\left({\varphi}\right)} \right ]_{\frac{\pi}{4}}^{\frac{\pi}{2}}d\theta$

V de

T V 33 4

$\text{V}=\dfrac{2\pi}{9}$