Thread: Volume of body obtained by rotating around the x axis and y axis.

Hi,

If y= 4x^3+3x^2-4x-3, calculate the volume of the body obtained by rotating y around 1) the x axis 2) the y axis.

I'm at a bit of a loss here. I know the formula for rotating about the x axis is the integral of pi.y^2 and integral of pi.x^2 for rotation about the y axis.

For the x axis rotation, i presume I have to square out the whole function and then integrate bit by bit. However, for rotation about the y axis, how do I express x^2 in terms of y? Bit confused, would appreciate help!

First, find the roots .

Hi.

I've factorised to get(x-1)(x+1)(4x+3). So the roots are x=1, x=-1 and x=-3/4

around the x-axis:

Thanks but do you know how to do it about the y axis? Can i say at y=0 that the above function squared then equals zero and then bring the x squared term over to the left and side...

Sorry, Just realised , can I use the shell method for rotating about the y axis? i.e that the volume of a solid rotated about the y axis can be written as the integral of 2pixf(x) between the limits a and b?

Originally Posted by orlacoon

Hi,

If y= 4x^3+3x^2-4x-3, calculate the volume of the body obtained by rotating y around 1) the x axis 2) the y axis.

I'm at a bit of a loss here. I know the formula for rotating about the x axis is the integral of pi.y^2 and integral of pi.x^2 for rotation about the y axis.

For the x axis rotation, i presume I have to square out the whole function and then integrate bit by bit. However, for rotation about the y axis, how do I express x^2 in terms of y? Bit confused, would appreciate help!

I see your problem. are you sure you have given us the complete question. For example it does not tell us between what values of x they are asking.