Dont understand the bounds of this integral, shell method.

[Agent M27]

Homework Statement

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

y=4x-x^{2} , y=4 , x=0

Homework Equations

V=2\pi\int p(x)h(x)

from a to b

The Attempt at a Solution

V=2\pi\int4x^{2}-x^{3}

The equation y=4x-x^2 is the equation for an upside down parabola with its vertex at (2,2) and has roots at x=0,4. When revolving around the y-axis the bounds are from a to b which in this case ought to be 0 to 4. Instead the book gives the same integral as I have found but with the bounds from 0 to 2, I have been banging my head against the wall on this one for too long, any ideas? Thanks in advance.

Joseph

 

[Quinzio]

Homework Statement

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

y=4x-x^{2} , y=4 , x=0

Homework Equations

V=2\pi\int p(x)h(x)

from a to b

The Attempt at a Solution

V=2\pi\int4x^{2}-x^{3}

The equation y=4x-x^2 is the equation for an upside down parabola with its vertex at (2,2) and has roots at x=0,4. When revolving around the y-axis the bounds are from a to b which in this case ought to be 0 to 4. Instead the book gives the same integral as I have found but with the bounds from 0 to 2, I have been banging my head against the wall on this one for too long, any ideas? Thanks in advance.

Joseph

You should pay attention to the given lines too.
y=4 , x=0

The area to revolve is bounded
From (0,0) to (0,4) along the x=0 axis.
From (0,4) to (2,4) along the y=4 axis.
From (2,4) to (0,0) along the parabola.

 

[Mark44]

Homework Statement

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

y=4x-x^{2} , y=4 , x=0

Homework Equations

V=2\pi\int p(x)h(x)

from a to b

The Attempt at a Solution

V=2\pi\int4x^{2}-x^{3}

The equation y=4x-x^2 is the equation for an upside down parabola with its vertex at (2,2) and has roots at x=0,4. When revolving around the y-axis the bounds are from a to b which in this case ought to be 0 to 4. Instead the book gives the same integral as I have found but with the bounds from 0 to 2, I have been banging my head against the wall on this one for too long, any ideas? Thanks in advance.

Joseph

I think you might be looking at the wrong region. The one that is described above is to the left of the parabola. It's a sort of triangular region bounded on the left by the y-axis (x = 0), above by the line y = 4, and on the right by the parabola. When you revolve it around the y axis, you get a sort of cone shape, with the vertex at the origin.

One thing for sure, your integrand doesn't take into account that h(x) extends from the parabola up to the line y = 4.

 

[Agent M27]

As I mentioned I have been staring at this one for to long my integral is actually:

V=2\pi\int x^{3}-4x^{2}+4x dx

From your direction Mark, they are not actually wanting me to find the volume within that upside down parabola, they want me to find the "open space" between it and the y-axis. I guess I can see this being that it gave me the other conditions of x=0 and y=4, but I still don't see why I would stop at x=2. If they had just provided the y=4x-x^{2} then they would be speaking about the parabola only, is that correct? Thanks.

Joe

 

[Mark44]

If the region being revolved had been bounded by y = 4x - x^2 and the x-axis, then yes, the region would be that part of the parabola.

As for why x = 2 is the upper limit of integration, where does the line y = 4 intersect the parabola?