Volume of Ellipse Cross Section Perpendicular to x-Axis

[JKLM]

If the area enclosed by an ellipse 4x^2+y^2=1 and its cross section is perpendicular to the x-axis then its volume is?

I don't have the slightest clue how to do this?

Maybe solve for 2y^2=1-4x^2 set the integral equal to pi times the intergral of 1/4 to 1 of 1-4x^2?

 

[Hurkyl]

It sounds to me like you're missing part of the problem...

 

[Nachtwolf]

Originally posted by JKLM
If the area enclosed by an ellipse 4x^2+y^2=1 and its cross section is perpendicular to the x-axis then its volume is?

I don't have the slightest clue how to do this?

Maybe solve for 2y^2=1-4x^2 set the integral equal to pi times the intergral of 1/4 to 1 of 1-4x^2?

My calculus is rusty but here's the first idea which comes to mind. Solve for y:

y = sqrt(1-4x^2)

Then do an integral with limits 0 to 1/4, and multiply the total by 4. In other words (using a clunky capital S as an integral sign), solve this:

4 [ S(0-.25) sqrt(1-4x^2) dx ]

And that should be the answer. You may have trouble solving such a weird integral, but give it a go and see if the answer is reasonable.


--Mark

 

[himanshu121]

How can a planar fig occupy a volume I agree with Hurkyl u are missing something it must be rotated about some axis to have a solid figure

 

[Nachtwolf]

In retrospect you're probably right (I wrote that response in a hurry). At the time I posted, I assumed he meant "area" since he provided insufficient information to solve a volume-related problem.


--Mark