MHB Volume of Solid with Circular Base & Equilateral Triangles

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SUMMARY

The volume of a solid with a circular base of radius 3 and equilateral triangle cross-sections perpendicular to the x-axis is calculated to be 36√3. The correct approach involves centering the circular base at the origin of the xy-plane and integrating the area of the triangular cross-sections above the first quadrant, then quadrupling the result due to symmetry. The formula used is V = 4 * (√3/2) * ∫(0 to 3) (9 - x²) dx. A common mistake is integrating from -3 to 3 and doubling the volume, which does not yield the correct result.

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  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the properties of equilateral triangles and their geometric relationships.
  • Knowledge of circular geometry and the equation of a circle.
  • Ability to perform definite integrals and apply symmetry in geometric problems.
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  • Study the derivation of volume formulas for solids of revolution.
  • Learn about the properties and applications of equilateral triangles in geometry.
  • Explore the use of integration in calculating areas and volumes of complex shapes.
  • Practice solving similar problems involving cross-sectional areas and volume calculations.
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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!
 
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