Find Volume of Solid Rotating About X-Axis: f(x)=4-x^2 0≤x≤2

[CompSciGuy]

Hello all, I have a problem that I'm not too sure if I'm doing correctly. Here it is, along with my work/answer thus far:

Given the function of the curve below, find the volume of the solid generated by rotating the area (A) about the x-axis.
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f[/FONT][FONT=&quot]([/FONT][FONT=&quot]x[/FONT][FONT=&quot])[/FONT][FONT=&quot]=4-x[/FONT][FONT=&quot]^[/FONT][FONT=&quot]2 0≤x≤2

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[Deleted member 4993]

Hello all, I have a problem that I'm not too sure if I'm doing correctly. Here it is, along with my work/answer thus far:

Given the function of the curve below, find the volume of the solid generated by rotating the area (A) about the x-axis.

f(x)=4-x^2 0≤x≤2

View attachment 8104

Please post a larger image - I cannot read it!

 

[HallsofIvy]

The function given is \(\displaystyle y= 4- x^2\) with x ranging from 0 to 2. I cannot read very well how you attempted it but this is what I would do:

Imagine dividing the area into many thin rectangles, with base length \(\displaystyle \Delta x\) and height \(\displaystyle y= 4- x^2\). Rotating around the x-axis, each such rectangle forms a disk with area \(\displaystyle \pi r^2= \pi (4- x^2)^2= \pi (16- 8x^2+ x^4)\) and thickness [/tex]\Delta x[/tex] so volume \(\displaystyle \pi (16- 8x^2+ x^4)\Delta x\). The total volume is given by the Riemann sum: \(\displaystyle \pi \sum (16- 8x^2+ x^4)\Delta x\). Taking the limit as the number of such rectangles becomes infinite, we have the integral \(\displaystyle \pi\int_0^2 (16- 9x^2+ x^4)dx\).

I can't read your final answer but it looks like you have \(\displaystyle 1\frac{3}{5}\). If so you have simply dropped the "\(\displaystyle \pi\)" at the last step!