Volume of Solid with Circular Base & Equilateral Triangles

[veronica1999]

A solid has a circular base of radius 3. If every plane cross section perpendicular to the x-axis is an equilateral triangle, then it's volume is

I keep on getting 18 root 3. But the answer is 36 root 3.

Could I get some help?

Thanks.

 

[MarkFL]

My approach would be to center the circular base at the origin of the $xy$-plane and then consider the volume above the first quadrant only, and then quadruple this volume given the symmetry of the object.

The cross-sections above the first quadrant are $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, and letting the base be $y$, we must then have the height as $\sqrt{3}y$, where:

\(\displaystyle y=\sqrt{9-x^2}\)

And so the total volume of the object is:

\(\displaystyle V=4\cdot\frac{\sqrt{3}}{2}\int_0^3 9-x^2\,dx\)

I think you will find you get the desired result upon integrating.

 

[veronica1999]

My approach would be to center the circular base at the origin of the $xy$-plane and then consider the volume above the first quadrant only, and then quadruple this volume given the symmetry of the object.

The cross-sections above the first quadrant are $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, and letting the base be $y$, we must then have the height as $\sqrt{3}y$, where:

\(\displaystyle y=\sqrt{9-x^2}\)

And so the total volume of the object is:

\(\displaystyle V=4\cdot\frac{\sqrt{3}}{2}\int_0^3 9-x^2\,dx\)

I think you will find you get the desired result upon integrating.

Thank you. That makes a lot of sense. The only thing I'm wondering now is why I kept getting the wrong answer.

I was trying to use a similar approach (I had the same integral but different intervals of integration). I integrated from -3 to 3 and doubled the volume instead. Why did this not work? Or did I make a careless mistake in calculating the final answer?

 

[MarkFL]

Thank you. That makes a lot of sense. The only thing I'm wondering now is why I kept getting the wrong answer.

I was trying to use a similar approach (I had the same integral but different intervals of integration). I integrated from -3 to 3 and doubled the volume instead. Why did this not work? Or did I make a careless mistake in calculating the final answer?

Can you show me your work? I will try to spot where the error might be. :D

 

[veronica1999]

Can you show me your work? I will try to spot where the error might be. :D

Actually I just realized my mistake. ^.^ I wasn't getting the correct anti-derivative. Thank you for your help!