Question
the curve f(x)=x^2 from x=0 to x=4 is revoloved around the axis x=5. Determine the volume of the solis of revolution using the method of curcular cross sections
Answers
Steve
Need more info. The curve is one boundary. Is the region to revolve bounded by y=0 and x=4, or by y=0 and x=0, or what?
Just revolving a curve will produce a surface, but with no corresponding area, there's no volume.
Just revolving a curve will produce a surface, but with no corresponding area, there's no volume.
Anonymous
That is all the information I was given, so that is why I am stuck on it.
Steve
Well, let's take a stab at it. The simplest region would be the region under the curve from x=0 to x=4, bonded by the line x=4. Kind of a little curved triangle.
If we rotate that around the line x=5 we'll have a kind of volcano-shaped solid with a hole down the center.
To find the volume using circular cross-sections, think of a stack of thin washers, getting smaller as they stack up.
Each washer has i hole inside of radius 1 (distance from x=4 to x=5).
The outside radius is 5-x.
Each washer has a thickness dy.
Now, y = x^2 so x = sqrt(y)
The area of each washer is pi*(5-x)^2 - pi*1^2 = pi(25 - 10x + x^2 - 1)
= pi(y - 10*sqrt(y) + 24)
The volume is Integral(pi*(y - 10y^1/2 + 24) dy [0,16]
That will be
pi (1/2 y^2 - 20/3 y^3/2 + 24y)[0,16]
= pi(128 - 1280/3 + 384)
= 256pi/3
If we rotate that around the line x=5 we'll have a kind of volcano-shaped solid with a hole down the center.
To find the volume using circular cross-sections, think of a stack of thin washers, getting smaller as they stack up.
Each washer has i hole inside of radius 1 (distance from x=4 to x=5).
The outside radius is 5-x.
Each washer has a thickness dy.
Now, y = x^2 so x = sqrt(y)
The area of each washer is pi*(5-x)^2 - pi*1^2 = pi(25 - 10x + x^2 - 1)
= pi(y - 10*sqrt(y) + 24)
The volume is Integral(pi*(y - 10y^1/2 + 24) dy [0,16]
That will be
pi (1/2 y^2 - 20/3 y^3/2 + 24y)[0,16]
= pi(128 - 1280/3 + 384)
= 256pi/3