Integration Volume: Finding the Rotation of Shaded Area

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SUMMARY

The discussion centers on calculating the volume of a solid generated by rotating the area between the curves y=x^2 and y=2-x^2 around the y-axis through π radians. The initial calculation yielded an incorrect volume of 16/15 π, while the correct volume is π. The user identified a mistake in their setup, realizing that the integration should be performed with respect to y rather than x, leading to the correct formulation of the volume integral.

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  • Understanding of calculus, specifically volume of revolution concepts.
  • Familiarity with integral calculus and the use of definite integrals.
  • Knowledge of the washer method for finding volumes of solids of revolution.
  • Ability to interpret and manipulate equations of curves.
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  • Review the washer method for calculating volumes of solids of revolution.
  • Practice setting up integrals for volumes with respect to both x and y.
  • Explore the concept of changing variables in integrals to simplify calculations.
  • Study examples of volume calculations involving multiple curves and their intersections.
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Students studying calculus, particularly those focusing on volume calculations in solid geometry, and educators looking for examples of common mistakes in integration setups.

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Homework Statement



Find the volume of solid generated when the shaded area is rotated through pi radians about the y-axis . The graphs are y=x^2 and y=2-x^2 .

Homework Equations





The Attempt at a Solution



i did everything and found the answer to be 16/15 pi but the answer given is pi .

[tex]V=\pi \int^{1}_{0}x^4 dx+\pi \int^{2}_{1}(2-x^2)^2 dx[/tex]

any mistakes in my set up ?
 
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When I set it up, I got:

[tex]V=\int_0^1\pi x\int_{x^2}^{2-x^2} dy dx = \int_0^1\pi x(2-2x^2)dx = 2\pi[\frac{x^2}{2}-\frac{x^4}{4}]_0^1 = \frac{\pi}{2}[/tex]

However, this gives pi/2, not pi. Are you sure pi is the right answer? Anyone else?
 
ok , i think i see my mistake , its with respect to y .
 

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