Question

find the volume of the sphere x^2+y^2+z^2=a^2 using triple integration​

Answer 1

Answer:

The equation of the cylinder is

x²+y²=ay⟹x²+(y−a2)2=a²4(1)

and the sphere is

x²+y²+z²=a²

In order to find this integral, I first realized that the region is a sphere inside a cylinder delocated from the center. I first computed the integral with respect to z of the length of the sphere, which is:

∫a−x²−y²√−a−x²−y²√dz=2a−x²−y²−−−−−−−−−√

Now, I need to take this integral over the circumference that the cylinder forms when projected to the xy plane, which is (1). Therefore, I need to integrate 2a−x²−y²−−−−−−−−−√ inside the circumference (1). I'll do it in polar coordinates, so I need to make my substitutions. The idea is to substitute x=pcos(t),y=psin(t) and not x=pcos(t),y−a2=psin(t) because I want to make the integrand easier, not the region. So the region will not be simply p from 0 to a2 and θ from 0 to 2π. If I make the substition x=pcos(t),y=psin(t) in (1) I end up with the equation p=sin(θ) for the circumference (1)in polar coordinates. I also know that θ must go from −π2 to π2 for this one. Therefore, my double integral becomes:

∫π2−π2∫sin(t)02a−(pcos(θ))²−(psin(θ))²−−−−−−−−−−−−−−−−−−−−−−√p dp dθ

Since this integral is quite hard, I would like to know if I'm doing all fine before trying to integrate this. Wolfram Alpha does not integrate it, don't know why, but my book says the answer is

2⋅a³π3

Answer 2

Answer:

see the attachment

Step-by-step explanation:

hope it helps you