Volume of Solid Generated by Ellipse Quadrant Revolving About Major/Minor Axis

[himanshu121]

The quadrant of the ellipse [tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}[/tex] = 1. lying in the first quadrant, revolves about the line joining the extremities of the major and minor axis. Show that the volume of the solid generated is [tex] \frac{\pi a^2 b^2}{\sqrt{a^2+b^2}}[/tex] ([tex]\frac{5}{3} - \frac{\pi}{2}[/tex]).

I tried the above question and it was a lot of intricated.Pls tell me some shortest route to the problem

 

[lethe]

this problem seems a little messed up, since the region to be rotated is on both sides of the axis of rotations. there will be overlap. maybe we should just choose that part of the ellipse that is in the first quadrant and above the line joining the extremities of the two axes?

 

[himanshu121]

I wrote the question correctly there might be some problems
But i think question requires the upper part which is above the line

 

[lethe]

Originally posted by himanshu121
I wrote the question correctly there might be some problems
But i think question requires the upper part which is above the line

i think so too, which is why i think there is a problem with what you wrote, which includes not just the part which is above the line, but instead the whole first quadrant of the ellipse.

are you sure that is correct?

 

[himanshu121]

It came in my exam and it says what i have written

 

[lethe]

Originally posted by himanshu121
It came in my exam and it says what i have written

ok, well i don t know what to do with it then...

sorry.

maybe you can ask your teacher if there is a typo...? or maybe someone else here can help you? the problem seems ambiguous to me.

 

[NateTG]

Originally posted by himanshu121
The quadrant of the ellipse [tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}[/tex] = 1. lying in the first quadrant, revolves about the line joining the extremities of the major and minor axis. Show that the volume of the solid generated is [tex] \frac{\pi a^2 b^2}{\sqrt{a^2+b^2}}[/tex] ([tex]\frac{5}{3} - \frac{\pi}{2}[/tex]).

I tried the above question and it was a lot of intricated.Pls tell me some shortest route to the problem

There are two extremes for each axis, so there are four possible axes of rotation. Perhaps you should ask your teacher which one he wanted?

 

[himanshu121]

What if we consider the revolution of triangle bounded by the line won't we come to conclusion as the upper curve by default would be included

 

[HallsofIvy]

NateTG: since the original problem referred to the first quadrant, I would think it reasonable to assume that the "extremities" referred to are (a,0) and (0,b). Then the axis of rotation is bx+ ay= ab.

The difficulty that lethe was referring to is that that axis goes through the figure. I would suspect that the problem was intended to be the figure generated if the boundary of the ellipse (above the axis of rotation) were rotated about that axis.