Integral to find the volume of a torus

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SUMMARY

The volume of the torus formed by revolving a circle of radius 2 centered at (3,0) about the y-axis can be calculated using Pappus' centroid theorem. The correct integral to evaluate is V = ∫_{-2}^2 π((3 + √(4 - y²))² - (3 - √(4 - y²))²) dy. This approach simplifies the calculation by considering the area of the circle and the distance traveled by its centroid. The final volume can be derived using trigonometric substitution for further simplification.

PREREQUISITES
  • Understanding of integral calculus and volume calculation
  • Familiarity with Pappus' centroid theorem
  • Knowledge of trigonometric substitution techniques
  • Ability to manipulate and evaluate definite integrals
NEXT STEPS
  • Study Pappus' centroid theorem in detail
  • Learn about trigonometric substitution methods in integral calculus
  • Practice evaluating definite integrals involving square roots
  • Explore applications of volume calculation in solid geometry
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Students studying calculus, particularly those focusing on volume calculations and geometric applications, as well as educators teaching integral calculus concepts.

togame
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Homework Statement


Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the y-axis. Use geometry to evaluate the integral.


Homework Equations


Formula for the semi-circle: y=\sqrt{4-x^2}
Solving for x give x=\pm\sqrt{4-y^2}


The Attempt at a Solution


I was thinking of solving the integral from -2 to 2 using both the positive and negative sides, then evaluate the integral and then multiply by 2, but I'm not so sure about my equation.

2\int_{-2}^2\pi\big( (3-\sqrt{4-y^2})^2 - (3+\sqrt{4-y^2})^2 \big) \mathrm d y

I just seem lost on this one :(
 
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togame said:

Homework Statement


Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the y-axis. Use geometry to evaluate the integral.

Homework Equations


Formula for the semi-circle: y=\sqrt{4-x^2}
Solving for x give x=\pm\sqrt{4-y^2}

The Attempt at a Solution


I was thinking of solving the integral from -2 to 2 using both the positive and negative sides, then evaluate the integral and then multiply by 2, but I'm not so sure about my equation.
2\int_{-2}^2\pi\big( (3-\sqrt{4-y^2})^2 - (3+\sqrt{4-y^2})^2 \big) \mathrm d y
I just seem lost on this one :(
I believe the equation is:

V=\int_{-2}^2\pi\big( (3+\sqrt{4-y^2})^2 - (3-\sqrt{4-y^2})^2 \big) \mathrm d y

V=\int_{-2}^2\pi\cdot 12(\sqrt{4-y^2}) \mathrm d y

then trigonometric substitution...
 
togame said:

Homework Statement


Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the y-axis. Use geometry to evaluate the integral.

Homework Equations


Formula for the semi-circle: y=\sqrt{4-x^2}
Solving for x give x=\pm\sqrt{4-y^2}

The Attempt at a Solution


I was thinking of solving the integral from -2 to 2 using both the positive and negative sides, then evaluate the integral and then multiply by 2, but I'm not so sure about my equation.

2\int_{-2}^2\pi\big( (3-\sqrt{4-y^2})^2 - (3+\sqrt{4-y^2})^2 \big) \mathrm d y

I just seem lost on this one :(

Edit : Got beat to it.
 
the way to do this is to use pappus' theorem. the product of the distance traveled by the center of the circle by the area of the circle.
 

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