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

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The discussion focuses on calculating the volume of a solid formed by revolving the area between the curve y=1-x^2 and the x-axis around the y-axis. The volume is determined using both the disk method and the shell method, yielding the same result. The disk method involves integrating the function to find the volume of disks, while the shell method calculates the volume of cylindrical shells. Both methods confirm that the volume of the solid is π/2. The calculations emphasize the importance of visualizing the region and applying appropriate integration techniques.
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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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