Finding the Volume of a Revolved Ellipse Using Calculus

[Smartass]

Homework Statement

The problem is that an ellipse (centered at origin) is revolved about y-axis. Now I have to find the volume of this swept region. But how do I go about using calculus? I have to derive it.

Homework Equations

Volume of ellipsoid = 4/3*pi*abc (source wikipedia)
Equation of ellipse: x^2/a^2 + y^2/b^2 = 1

The Attempt at a Solution

integral(pi*a*b*dθ), limits 0 to pi. What is wrong with this?


Now I know after searching that the volume is given by integral(pi*y^2*dx), limits -b to b, but how do you get this?

 

[HallsofIvy]

Homework Statement

The problem is that an ellipse (centered at origin) is revolved about y-axis. Now I have to find the volume of this swept region. But how do I go about using calculus? I have to derive it.

Homework Equations

Volume of ellipsoid = 4/3*pi*abc (source wikipedia)
Equation of ellipse: x^2/a^2 + y^2/b^2 = 1

The Attempt at a Solution

integral(pi*a*b*dθ), limits 0 to pi. What is wrong with this?

What's right with it? Why would you expect it to give you the volume?
That would, of course, give \pi^2 ab. If a= b= r, your ellipsoid will be a sphere of radius r. Does \pi^2 r^2 give the volume of that sphere?

Now I know after searching that the volume is given by integral(pi*y^2*dx), limits -b to b, but how do you get this?

Actually, that's wrong. If the ellipse goes from x=-a to x= a, it wouldn't make sense to integrate dx from -b to b. If you were to integrate from -a to a, you would get the volume if that same ellipse were rotated around the x-axis. The correct formula is
\pi\int_{y=-b}^b x^2 dy

If you take a cross section of the rotated ellipsoid, perpendicular to the y-axis (axis of rotation) you get a circle of radius "x". It has area \pi x^2 and, imagining it as a thin disk of thickness \Delta y (the thickness is measured along the y-axis), volume \pi x^2 \Delta y. Summing over all such disks gives you a Riemann sum approximating the true volume. In the limit, the Riemann sums become the integral \int_{y=-b}^b \pi x^2 dy.

The real reason for learning about Riemann sums (you certainly never use them to actually calculate an integral!) is because most integrals are set up using them.

 

[Smartass]

What's right with it? Why would you expect it to give you the volume?
That would, of course, give \pi^2 ab. If a= b= r, your ellipsoid will be a sphere of radius r. Does \pi^2 r^2 give the volume of that sphere?



Actually, that's wrong. If the ellipse goes from x=-a to x= a, it wouldn't make sense to integrate dx from -b to b. If you were to integrate from -a to a, you would get the volume if that same ellipse were rotated around the x-axis. The correct formula is
\pi\int_{y=-b}^b x^2 dy

If you take a cross section of the rotated ellipsoid, perpendicular to the y-axis (axis of rotation) you get a circle of radius "x". It has area \pi x^2 and, imagining it as a thin disk of thickness \Delta y (the thickness is measured along the y-axis), volume \pi x^2 \Delta y. Summing over all such disks gives you a Riemann sum approximating the true volume. In the limit, the Riemann sums become the integral \int_{y=-b}^b \pi x^2 dy.

The real reason for learning about Riemann sums (you certainly never use them to actually calculate an integral!) is because most integrals are set up using them.

That makes so much sense, thanks! To justify my attemp, what I did was evaluate the area of the ellipse (=pi*a*a) and then if I rotate it by a small angle, say dθ, then region swept by it would be the volume of that region. If I take dθ from 0 to pi, then one entire rotation of the ellipse takes place and volume of the entire region. I knew I was wrong, it just didn't feel right. But what you said is real good!