Aug 11, 2017 · The volume of sphere using integrals Ask Question Asked 8 years, 3 months ago Modified 1 year, 9 months ago

Sep 2, 2020 · I was asked to explain why the volume of a sphere is $\\frac{4}{3}\\pi r^3$ to a student that does not have the knowledge of calculus. In doing so I thought of an argument and I cannot seem to …

2 With given radius r of a sphere let the inscribed cone have height h then remaining length without radius is (h-r) let R be radius of cone then there we get a right angle triangle with r as hypotanious R …

Dec 21, 2019 · Find the volume of the solid within the sphere $x^2+y^2+z^2=9$, outside the cone $z=\sqrt {x^2+y^2}$, and above the $xy$-plane. Using Cylindrical coordinates, $r^2+z^2=9$ and $z=r$.

Jul 20, 2021 · I actually have found the solution using double integral in polar coordinate. However, I am curious about whether I could find the same exact solution using triple integral in spherical …

You can see that not all such cylinders have equal volume, just by considering the extreme case of when two of the points are very close together. You get either a long thin rod, or a big flat pancake, …

Jun 8, 2019 · A sphere is a 3-dimensional object. The 2-dimensional analogue of a sphere is a circle.

Jan 23, 2021 · If this is the volume of the sphere, I am not sure how I am supposed to show this using the shell method. Any assistance would help! A picture below depicts the problem.

Jan 18, 2020 · Thank you very much. It is indeed heartening to see that limit from $-\pi/2$ to $\pi/2$ also produces the same result. But, could you please clarify a thing for me. When I calculated the …

Also, the upper bound for $\rho$ should have been $6\cos\phi$ not $3$. The equation $\rho = 3$ describes a sphere of radius $3$ centered at the origin while the equation $\rho = 6\cos\phi$ …