Question

find the volume bounded by x²+y²+z²=9 using triple integration.​

Answer 1

SOLUTION

TO DETERMINE

The volume bounded by x² + y² + z² = 9 using triple integration

EVALUATION

Here the given equation of the sphere is

x² + y² + z² = 9

Comparing with the general equation

x² + y² + z² = a² we get

a² = 9

⇒ a = 3

Changing to polar spherical coordinates by putting

x = r sin θ cos Φ , y = r sin θ sin Φ , z = r cos θ

We have dx dy dz = r² sin θ dr dθ dΦ

Also the volume of the sphere is 8 times the volume of its portion in the positive octant for which r varies from 0 to 3 , θ varies from 0 to π/2 and Φ varies from 0 to π/2

∴ Volume of the sphere

$\displaystyle \sf{ = 8 \: \int\limits_{0}^{3} \int\limits_{0}^{ \frac{\pi}{2} } \int\limits_{0}^{ \frac{\pi}{2}} } \: {r}^{2} \sin \theta \, \: dr \: d \theta \: d \phi$

$\displaystyle \sf{ = 8 \: \int\limits_{0}^{3} {r}^{2} \: dr \: \int\limits_{0}^{ \frac{\pi}{2} }\sin \theta \:d \theta \int\limits_{0}^{ \frac{\pi}{2}} } \: \: d \phi$

$\displaystyle \sf{ = 8 \times \frac{ {r}^{3} }{3} \bigg|_0^3 \times - \cos \theta \bigg|_0^{ \frac{\pi}{2} } \times \: \phi \bigg|_0^{ \frac{\pi}{2} }}$

$\displaystyle \sf{ = 8 \times \bigg( \frac{ {3}^{3} }{3} - 0 \bigg) \times \bigg(- \cos \frac{\pi}{2} + \cos 0 \bigg) \times \bigg( \frac{\pi}{2} - 0 \bigg)}$

$\displaystyle \sf{ = 8 \times 9 \times 1 \times \frac{\pi}{2} }$

$\displaystyle \sf{ = 36\pi }$

FINAL ANSWER

Volume of the sphere = 36π cubic unit

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. If integral of (x^-3)5^(1/x^2)dx=k5^(1/x^2), then k is equal to?

https://brainly.in/question/15427882

2. If integral of (x^-3)5^(1/x^2)dx=k5^(1/x^2), then k is equal to?

https://brainly.in/question/15427882