Huzafa's question at Yahoo Answers regarding a solid of revolution

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SUMMARY

The volume of the solid obtained by revolving the area between the graph of y=1-x² and the x-axis around the y-axis is calculated using both the disk and shell methods. The disk method yields a volume of V=π/2, derived from the integral V=π∫₀¹(1-y)dy. Similarly, the shell method also confirms the volume as V=π/2, calculated through the integral V=2π∫₀¹(x-x³)dx. Both methods validate the result, demonstrating the consistency of calculus techniques in finding volumes of solids of revolution.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the disk and shell methods for calculating volumes of solids of revolution.
  • Knowledge of graphing functions and interpreting areas under curves.
  • Ability to manipulate and integrate polynomial functions.
NEXT STEPS
  • Study the disk method for solids of revolution in more complex scenarios.
  • Explore the shell method in detail, including its applications in different coordinate systems.
  • Learn about the implications of even and odd functions in volume calculations.
  • Investigate other methods for finding volumes, such as the washer method.
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Students and professionals in mathematics, particularly those studying calculus, as well as educators seeking to enhance their understanding of volume calculations for solids of revolution.

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Here is the question:

Find volume of solid obtained by revolving around y-axis the plane area btw the graph y=1-x^2 and the x-axis?

I have posted a link there to this topic so the OP can see my work.
 
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Hello huzafa,

The first thing I would do is draw a diagram of the region to be revolved. We need only concern ourselves with either the quadrant I area or the quadrant II area because we are revolving an even function about the $y$-axis. I will choose to plot the quadrant I area:

View attachment 974

Using the disk method, we observe that the volume of an arbitrary disk is:

$$dV=\pi r^2\,dy$$

where:

$$r=x\,\therefore\,r^2=x^2=1-y$$

and so we have:

$$dV=\pi(1-y)\,dy$$

Summing the disks by integration, we have:

$$V=\pi\int_0^1 1-y\,dy=\pi\int_0^1 u\,du=\frac{\pi}{2}\left[u^2 \right]_0^1=\frac{\pi}{2}$$

Using the shell method, we observe that the volume of an arbitrary shell is:

$$dV=2\pi rh\,dx$$

where:

$$r=x$$

$$h=y=1-x^2$$

and so we have:

$$dV=2\pi\left(x-x^3 \right)\,dx$$

Summing the shells by integration, we find:

$$V=2\pi\int_0^2 x-x^3\,dx=\frac{\pi}{2}\left[2x^2-x^4 \right]_0^1=\frac{\pi}{2}$$
 

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