Use spherical coordinates. evaluate (x2 + y2) dv e , where e lies between the spheres x2 + y2 + z2 = 1 and x2 + y2 + z2 = 25.

To solve this problem, it is necessary determine the integrations limits first.

The solution is:

V = 1666,13×π vol units

In order to express the information in spherical coordinates, we have:

x = ρ×cosθ×sinφ

y = ρ×sinθ×sinφ

z = ρ×cosφ

and     dV = ρ²×sinφ×dρ×dθ×dφ

0 ≤ φ ≤ π

0 ≤ θ ≤ 2×π

x² + y² + z² = 25       and         x² + y² + z² = 1    are two concentric spheres with center at the origin and radius 5 and 1 respectevely .

According to that:

1 ≤  ρ ≤ 5

V = ∫∫∫ ( x² + y² )× dV  

V  =   ∫∫∫ ( x² + y² )×ρ²×sinφ×dρ×dθ×dφ

V =  ∫∫∫ (  ρ²×cos²θ×sin²φ  + ρ²×sin²θ×sin²φ ) ×ρ²×sinφ×dρ×dθ×dφ

V =  ∫∫∫ [  ρ²×sin²φ ( cos²θ + sin²θ )  ×ρ²×sinφ×dρ×dθ×dφ

V =  ∫∫∫ [  ρ²×sin²φ×ρ²×sinφ×dρ×dθ×dφ

V =  ∫∫∫ [ ρ⁴×sin³φ×dρ×dθ×dφ

V =  ∫₁⁵ρ⁴×dρ ×∫dθ ×∫sin³φ×dφ

V =  ρ⁵/5 |₁⁵×∫dθ ×∫sin³φ×dφ

ρ⁵/5 |₁⁵  = 5⁵/5 - 1⁵/5   = 3125 / 5 - 1/5  = 3124/5

V = 3124/5× θ |₀²π×∫sin³φ×dφ

θ |₀²π  = 2×π - 0

V = 3124/5× (2×π) ×∫sin³φ×dφ

∫sin³φ×dφ  [ 0  y π ] = 4/3   (from Simbolab)

V = 3124/5× (2×π) ×4/3

V = 1666,13×π   vol units

Related Lnk :link

the answer is c

hope this !