valtricotinevh

2022-07-17

Conceptual doubt in taking elements for integration
In physics, we were taught to find the center of mass using calculus for different shapes. I tried to find the volume of a sphere using similar concepts.
I took a cylindrical element to integrate whose radius is $R\mathrm{sin}\theta$ where $\theta$ is the angle made with the vertical to the edge of the element . And took the height to be $Rd\theta$ when I integrated all the cylindrical elements varying theta from o, $\pi$ and did not get the right answer.
I realize the error is taking $Rd\theta$ as height but why is it that we can take the "extension" to be $rd\theta$ when finding the surface area of the sphere.
To state my question more clearly:
Why cant the height of the element be $Rd\theta$ when finding volume since it works when finding surface area.
What I did to find volume:
${\int }_{0}^{\pi }\pi \left(R\mathrm{sin}\theta {\right)}^{2}\left(Rd\theta \right)$
What I did to find surface area:
${\int }_{0}^{\pi }2\pi R\mathrm{sin}\theta \left(Rd\theta \right)$

eri1ti0m

Expert

Step 1
The difference in the method of integration is that in the case of area, you need to measure a slant distance along the surface of the sphere from one circular edge of the area element to the other, whereas in the case of volume you need to measure the perpendicular distance from one circular face of the cylinder to the other.
The differential $R\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta$ accurately represents the slant distance along the surface of the sphere, which is perfect for measuring the surface area, but the surface of the sphere is not perpendicular to the circular faces of the cylindrical element (except when $\theta =\frac{\pi }{2}$), and therefore $R\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta$ is not generally an accurate representation of the height of the cylindrical element.
The plane that cuts the sphere at an angle $\theta$ from the axis is at a distance $R\mathrm{cos}\left(\theta \right)$ from the center. If you increase the angle to $\theta +\mathrm{\Delta }\theta$ then the distance of the plane from the center is $R\mathrm{cos}\left(\theta +\mathrm{\Delta }\theta \right)$. The cylindrical element for the change in angle $\mathrm{\Delta }\theta$ is between these two planes, so the height of the cylindrical element is $|R\mathrm{cos}\left(\theta +\mathrm{\Delta }\theta \right)-R\mathrm{cos}\left(\theta \right)|$.
Step 2
To use this in an integral, take the limit
$\underset{\theta \to 0}{lim}\frac{|R\mathrm{cos}\left(\theta +\mathrm{\Delta }\theta \right)-R\mathrm{cos}\left(\theta \right)|}{\mathrm{\Delta }\theta }=R\mathrm{sin}\theta ,$, which tells you that the perpendicular distance between circular cross-sections is represented by $R\mathrm{sin}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta .$.
So the correct integral for the volume is ${\int }_{0}^{\pi }\pi \left(R\mathrm{sin}\theta {\right)}^{2}\left(R\mathrm{sin}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta \right).$.
If you still do not see how to correctly translate the angle $\theta$ into the height of a cylindrical cross-section, try integrating by a variable that translates directly into the height of each cylinder. Put the sphere at the origin of a Cartesian system of coordinates and integrate along the z axis. The radius of the cylindrical element for any given value of z is $\sqrt{{R}^{2}-{z}^{2}}$ and the height of the element is dz, so the integral is
${\int }_{-R}^{R}\sqrt{{R}^{2}-{z}^{2}}\mathrm{d}z.$
Make the trig substitution $z=-R\mathrm{cos}\theta$ to get an integral in terms of $\theta .$. Note that in this substitution, you will replace dz with $R\mathrm{sin}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta .$.

Do you have a similar question?