How Do You Calculate the Volume of a Solid Revolved Around the Y-Axis?

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SUMMARY

The volume of a solid revolved around the Y-axis, defined by the area under the curve e^x - 1 and bounded by y=1, x=0, and x=ln2, can be calculated using the shell method. The correct integral setup is 2π∫₀^(ln2)x(1 - (e^x - 1))dx. The integration process involves using integration by parts, leading to the expression e^x(x - 1) - x² evaluated from 0 to ln2. The final volume is confirmed to be 2π(ln2 - 1)², correcting the initial miscalculation of constants during integration.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the shell method for calculating volumes of solids of revolution.
  • Knowledge of exponential functions and their properties.
  • Ability to perform integration by parts.
NEXT STEPS
  • Study the shell method in detail for calculating volumes of solids of revolution.
  • Practice integration by parts with various functions, focusing on exponential functions.
  • Explore advanced topics in calculus, such as multiple integrals and their applications in volume calculations.
  • Review common mistakes in integration to avoid errors in future calculations.
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Students and professionals in mathematics, particularly those studying calculus and volume calculations, as well as educators teaching integration techniques and solid geometry.

James889
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Hi,
The area
e^x-1
Is rotated about the y axis, bounded by y=1, x=0 and x=ln2 find the volume of the solid.

And i am clearly making something wrong, so if anyone could verify my work.

~ 2\pi\int_0^{ln2}x\cdot(1-(e^x-1)
-2\pi\int_0^{ln2}xe^x-2x

Integration:
u=x, du=1
dv=e^x, v=e^x

\int xe^x -2x= xe^x -e^x -x^2
e^x(x-1)-x^2\bigg|_0^{ln2} = 2\pi \cdot ln2-1 -(ln2)^2

The answer is supposed to be 2\pi(ln2-1)^2
Thanks!
 
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where did that 1 come from in your shell method formula?
 
When you integrate you should obtain : 2*pi*(ln22-2*ln2+1)

That then can be the factored to the book's answer.

You lost a 2 with the ln2
 

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