Why Is My Calculus Volume Integration Result Incorrect?

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SUMMARY

The discussion centers on calculating the volume of a solid of revolution formed by rotating the area between the curves y=x²+1 and y=9-x² around the x-axis. The correct volume is established as 640(pi)/3, while the user repeatedly calculates 896(pi)/3 due to algebraic errors in expanding the integrals. The integration setup involves using the formula 2π∫[(9-x²)² - (x²+1)²] dx, and the user is advised to carefully check their algebraic expansions to resolve the discrepancy.

PREREQUISITES
  • Understanding of solid of revolution concepts in calculus
  • Proficiency in performing definite integrals
  • Familiarity with algebraic manipulation and polynomial expansion
  • Knowledge of the disk method for volume calculation
NEXT STEPS
  • Review the disk method for calculating volumes of solids of revolution
  • Practice algebraic expansion of polynomials to avoid common mistakes
  • Learn about the washer method for volumes when dealing with two curves
  • Explore integration techniques in calculus, focusing on definite integrals
USEFUL FOR

Students studying calculus, particularly those focusing on volume calculations, educators teaching solid of revolution concepts, and anyone looking to improve their integration skills.

kira137
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I've been trying this equation for few hours now and I can't seem to get the right answer..
answer is suppose to be 640(pi)/3

Homework Statement

Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axisThe attempt at a solution
``2
Pi S (9-x^2)^2 - (x^2+1)^2 dx
``-2

``2
Pi S (81-2x^2+x^4-x^4-2x^2-1) dx
``-2

Pi[80x - (4x^3)/3] (x=-2 to 2)

= (448(pi)/3) - (-448(pi)/3) = 896(pi)/3

that's what I keep getting..
and if that isn't the correct answer for rotating around x=0, isn't it suppose to be the answer for when rotating around y=1?

thank you in advance
 
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kira137 said:
I've been trying this equation for few hours now and I can't seem to get the right answer..
answer is suppose to be 640(pi)/3

Homework Statement

Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axis


The attempt at a solution
``2
Pi S (9-x^2)^2 - (x^2+1)^2 dx
``-2

``2
Pi S (81-2x^2+x^4-x^4-2x^2-1) dx
``-2

Pi[80x - (4x^3)/3] (x=-2 to 2)

= (448(pi)/3) - (-448(pi)/3) = 896(pi)/3

that's what I keep getting..
and if that isn't the correct answer for rotating around x=0, isn't it suppose to be the answer for when rotating around y=1?

thank you in advance

Check again your algebra when you expand the squares inside the integration.

Cheers -- sylas
 

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