Volume of Solid of Revolution with Disk Method: y=5x^2, x=1, y=0 about x-axis

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SUMMARY

The volume of the solid of revolution formed by rotating the region bounded by the curves y=5x², x=1, and y=0 about the x-axis is calculated using the disk method. The correct volume is derived from the integral of πy² dx, where y=5x². The integral should be set up as π∫(0 to 1)(5x²)² dx, leading to the correct evaluation of π∫(0 to 1)(25x^4) dx. The final volume is (25π/5) = 5π, confirming the importance of correctly applying the disk method.

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  • Understanding of calculus, specifically integration techniques.
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  • Ability to evaluate definite integrals.
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I was wondering if anyone could help me with this. I'm stuck and not sure where to start/how to go about it and finding the integral as well...

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=5x^2,x=1,y=0, about the x-axis
 
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Imagine a line drawn perpendicular to the x-axis up to the give parabola. Rotated around the x-axis, it will sweep out a disk with area \pi y^2 so, imagining a thickness of "dx", volume \pi y^2 dx. Integrate that from 0 to 1.
 
HallsofIvy said:
Imagine a line drawn perpendicular to the x-axis up to the give parabola. Rotated around the x-axis, it will sweep out a disk with area \pi y^2 so, imagining a thickness of "dx", volume \pi y^2 dx. Integrate that from 0 to 1.
so when I do that.. I come up with (pi)(int 0-->1)(5*((x^3)/3)= (pi) (int. 0-->1) 5(1^3)/3= 5pi/3. But it says this answer is incorrect. Where am I going wrong?
 
If y= 5x^2 then y^2 is NOT equal to "5x^3/3"!

Or did you mean that 5x^3/3 is the result of integrating x^2 dx? (If so do NOT continue to write "int. 0-->1"!)

That would be the integral of \pi x^2dx but that is NOT what you should be integrating! You should be integrating \pi y^2 dx. And, of course, y= 5x^2.
 
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I think the disk method in terms of x should work.
 

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