Triple Integrals: Spherical Coordinates: sqrt(x^2 + y^2 + z^2 )

Today we will talk about calculus. Today’s topic is about spherical coordinates. Now you will see the technique. So we have x squared plus y squared plus z squared equals to p squared. Next is dzdxdy equals p squared sin Ø dpd Ø d theta. Now we need to change the boundary. It should be about from zero to number three. Let's say the radius of the circle is from zero to pi and from zero to 2 pi. So we get p, p squared sin Ø dp d Ø d theta. Now we just calculate this one. So p multiplied by p squared and you get p to the power of 3. And we get p to the power of 4 over 4, and we have sin Ø. We have the boundary from zero to number three. So what we end up with is 81 over four. Then we have eighty one over four negative courtside, and we have the boundary from zero to pi. So what we get is 2pi and eighty one over two d theta. The last one is we’ve got 81 over two multiplied by 2pi minus zero. So we get 81pi, and this is the final answer.

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