Question

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

Answer 1

The figure is symmetric, with equal volume in each of the eight octants, so we focus on the first octant, and multiply by 8. Let's work in cylindrical coordinates. In that case, our function is z=a2−r2−−−−−−√, and our region of integration is bounded by

0≤θ≤π20≤r≤acosθ

The volume is then:

8∫π20∫acosθ0∫a2−r2√0rdzdrdθ=8∫π20∫acosθ0ra2−r2−−−−−−√drdθ=2a3π3

Recall that the volume element in cylindrical coordinates is rdzdrdθ. The above integral is then evaluated by the substitution x=a2−r2.

Notice that our answer is half of the volume of the sphere, so there is an equal volume inside of the cylinder bounded by the sphere as there is outside of the cylinder bounded by the sphere.

Answer 2

Answer:

Volume of the sphere = $\frac{4}{3}\pi a^3$

Step-by-step explanation:

We are given the equation of a sphere

To Find : Area of the sphere using triple integration.

Solution :

Given Sphere equation :

$x^2+y^2+z^2=a^2$

Solving for z we get

$-\sqrt{a^2-y^2-x^2}\leq z \leq \sqrt{a^2-y^2-x^2}$

Now solving for y when z = 0

$-\sqrt{a^2-x^2} \leq y \leq \sqrt{a^2-x^2$

Finally limit of x is as follows

$-a \leq x \leq a$

we know that the Volume of the sphere

$\text{Volume}=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\int_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}} dxdydz$

$\text{Volume}=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}2{\sqrt{a^2-x^2-y^2}} dxdy$

$\text{Volume}=\int_{-a}^{a}2\frac{1}{2}\pi (a^2-x^2)dx$

           = $\pi (2a^3-\frac{2}{3}a^3)=\frac{4}{3}\pi a^3$

Hence we have calculated that the area of the sphere =  $\frac{4}{3}\pi a^3$

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