MHB Volume of Solid with Circular Base & Equilateral Triangles

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The discussion revolves around calculating the volume of a solid with a circular base of radius 3, where each cross-section perpendicular to the x-axis is an equilateral triangle. One participant initially calculated the volume as 18√3 but later realized the correct answer is 36√3. They centered the circular base at the origin and focused on the first quadrant, planning to quadruple the volume due to symmetry. The confusion arose from integrating over the wrong intervals and not obtaining the correct anti-derivative. Ultimately, the correct approach involves integrating from 0 to 3 and properly applying the formula for the volume of the solid.
veronica1999
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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.
 
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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:

$$y=\sqrt{9-x^2}$$

And so the total volume of the object is:

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

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

$$y=\sqrt{9-x^2}$$

And so the total volume of the object is:

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

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?
 
veronica1999 said:
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
 
MarkFL said:
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!