The discussion revolves around finding the volume of a solid defined by the equations of a sphere and a cylinder, specifically the region inside both \( x^2 + y^2 + z^2 = 4 \) and \( x^2 + y^2 = 1 \). Participants explore the implications of the term "inside" and how it affects the interpretation of the problem.
The conversation is ongoing, with participants offering different interpretations and methods for calculating the volume. Some express concerns about the correctness of others' approaches, while others emphasize the importance of visualizing the region of integration to avoid errors in setting up limits.
There are mentions of the complexity of the problem, with some participants suggesting that it may be more suitable for undergraduate students rather than advanced learners. The discussion also touches on the necessity of justifying methods and critically assessing calculations.
## z = \sqrt{4-r^2} ## and ## \theta = 2\pi r## where r= radius of sphere =2 ## \therefore \theta = 4\pi##Delta2 said: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.
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.WMDhamnekar said:## z = \sqrt{4-r^2} ## and ## \theta = 2\pi r## where r= radius of sphere =2 ## \therefore \theta = 4\pi##
This does work.WMDhamnekar said:## \displaystyle\int_{r=0}^1 4\pi r\sqrt{4-r^2} dr = \frac{\pi(32- 12\sqrt{3})}{3} \sim 11.745 ##
##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.WMDhamnekar said:## z = \sqrt{4-r^2} ## and ## \theta = 2\pi r## where r= radius of sphere =2 ## \therefore \theta = 4\pi##
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 2nuuskur said: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?
The answer to my question might become apparent if we slightly rewrite the integral asWMDhamnekar said:Sorry. I don't know the answer to your question.
If we convert from one coordinate system to another, then the Jacobian becomes relevant. Not relevant for method 2.WMDhamnekar said:By the way, I also want to know if the jacobian have been used in this Corrected Method 2