Change of variables in multiple integrals

[WMDhamnekar]

I know you didn't ask for my comment but it might be another method with "hidden" justification, just like yours is. You have to provide more details (at least to me) on why the integrand is $4\pi r\sqrt{4-r^2}$ and also why the limits of integration are from 0 to 1.

$z = \sqrt{4-r^2}$ and $\theta = 2\pi r$ where r= radius of sphere =2 $\therefore \theta = 4\pi$

 

[Delta2]

$z = \sqrt{4-r^2}$ and $\theta = 2\pi r$ where r= radius of sphere =2 $\therefore \theta = 4\pi$

Sorry your answer is too laconic for my standards , in order to understand. Why we take that $z$, why that $\theta$ and why we multiply them? Also r seems to vary from 0 to 1 not from 0 to 2.

 

[nuuskur]

$\displaystyle\int_{r=0}^1 4\pi r\sqrt{4-r^2} dr = \frac{\pi(32- 12\sqrt{3})}{3} \sim 11.745$

This does work.

$z = \sqrt{4-r^2}$ and $\theta = 2\pi r$ where r= radius of sphere =2 $\therefore \theta = 4\pi$

$2\pi r$ is the perimeter of a circle with radius $r$. Multiplied by some "height", we get the surface area of the side of a cylinder. The intuition here is that the volume of the (upper/lower half of the) body is obtained by summing up infinitely many cylinders with appropriate heights.

Here's my question to you. Why does the integral, as given, account for the entire volume and not just one of the halves?

 

[WMDhamnekar]

This does work.

$2\pi r$ is the perimeter of a circle with radius $r$. Multiplied by some "height", we get the surface area of the side of a cylinder. The intuition here is that the volume of the (upper/lower half of the) body is obtained by summing up infinitely many cylinders with appropriate heights.

Here's my question to you. Why does the integral, as given, account for the entire volume and not just one of the halves?

Sorry. I don't know the answer to your question. If you answer it yourself, it would be better for me and other readers, viewers, mathematics students and members of this Physics Forum. By the way, I also want to know if the jacobian have been used in this Corrected Method 2

 

[nuuskur]

Sorry. I don't know the answer to your question.

The answer to my question might become apparent if we slightly rewrite the integral as
<br /> V = 2\int _0^1 2\pi r\sqrt{4-r^2}\mathrm{d}r.<br />

By the way, I also want to know if the jacobian have been used in this Corrected Method 2

If we convert from one coordinate system to another, then the Jacobian becomes relevant. Not relevant for method 2.