Solids of Revolution, the negatives don't matter right?

[flyingpig]

Homework Statement

Use the method of cylindrical shells to nd the volume of the solid obtained by rotating
the region bounded by the curves y = x⁴, y = 1, x = 0 about the y-axis. Sketch the region and a typical shell

The Attempt at a Solution

I am just going to set it up, but I am not going to evaluate it

[tex]V = 2\pi \int_{0}^{1} x (x^4) dx[/tex]

My friend thinks it should be

[tex]V = 2\pi \int_{-1}^{1} x (x^4) dx[/tex]

Mine yields [tex]\frac{\pi}{3}[/tex] and my friend yields [tex]\frac{2\pi}{3}[/tex]

My argument is that we are only rotating the first quadrant. If we take the negative quadrant, it's meaningless. I am not even sure what we get if we take the negative quadrant, does it make it thicker?

EDIT: It is y = x⁴, not x²

 

[Mark44]

Homework Statement

Use the method of cylindrical shells to nd the volume of the solid obtained by rotating
the region bounded by the curves y = x², y = 1, x = 0 about the y-axis. Sketch the region and a typical shell

The Attempt at a Solution

I am just going to set it up, but I am not going to evaluate it

[tex]V = 2\pi \int_{0}^{1} x (x^4) dx[/tex]

My friend thinks it should be

[tex]V = 2\pi \int_{-1}^{1} x (x^4) dx[/tex]

Mine yields [tex]\frac{\pi}{3}[/tex] and my friend yields [tex]\frac{2\pi}{3}[/tex]

My argument is that we are only rotating the first quadrant. If we take the negative quadrant, it's meaningless. I am not even sure what we get if we take the negative quadrant, does it make it thicker?

Both setups are wrong. Using shells, your limits of integration are correct, but neither of you has the right integrand. A typical volume element is [itex]\Delta V = 2 \pi x (1 - x^2) \Delta x[/itex]

[itex]\Delta x[/itex] ranges from 0 to 1.

Did either of your graph the region?

 

[SammyS]

Neither is correct.

The side of each cylinder goes from y = x² to y = 1, therefore the height of each is: 1-x². The circumference of each is 2πx, so the volume element is:

dV = 2πx(1-x²)dx .

But you're right in the the limits of integration are from x=0 to x=1.

 

[flyingpig]

Okay guys, I made a mistake, the function is x² and not x⁴

 

[flyingpig]

Okay you know what I just figured out what went wrong. Lol thanks guys