Question

An ellipse whose major axis is $2a$ and the minor axis is $2b$ revolves around the minor axis. Then, find the volume of the solid that was created due to the rotation.

Answer 1

In the problem it is given an ellipse. So, we are able to use any kind of ellipse. Circle is also an ellipse in which lengths of both the axes will be equal. So, by comparing the formulas of the circle we know, we can easily get that for the ellipse too.

Complete step-by-step answer:
This kind of problem looks crazy and we can solve them in multiple ways. As per the data given in the question, there is an ellipse whose major axis is $2a$ and the minor axis $2b$ revolves around the minor axis.
So, we can imagine any kind of ellipse and all of us know that the circle Is also an ellipse in which the length of the major axis is equal to the length of the minor axis.
Imagine we rotate the circle around the diameter then we get the sphere.
We know that the volume of the sphere is $\dfrac{{4\pi {r^2}}}{3}$ where $r$ is the radius of the circle.
In the similar way,
If we rotate the ellipse around any of its axis,
Then the volume of the obtained ellipsoid can be either $\dfrac{{4\pi {a^2}b}}{3}$ or $\dfrac{{4\pi a{b^2}}}{3}$ according to the axis through which it is rotated.
Here $a$ refers to the length of semi major axis and $b$ refers to the length of semi minor axis.
So, the value of $a$ is always greater than $b$.
And as per the data given in the question, the ellipse is rotated around its minor axis.
The ellipsoid that formed now will have larger volume as compared to the ellipsoid formed when it is rotated around the major axis because of its structure.
In the possible values of $\dfrac{{4\pi {a^2}b}}{3}$ and $\dfrac{{4\pi a{b^2}}}{3}$, $\dfrac{{4\pi {a^2}b}}{3}$ will be the larger one because we know that a is larger than the b.
Hence the volume of the ellipsoid formed is $\dfrac{{4\pi {a^2}b}}{3}$

Note: These are some of the tricky type problems but can be solved very logically. Always try to search for the easiest way possible to solve this kind of problem. Compare the structures and analyze them through imagination. You can also able to do this by using differential calculus which may look quite higher.