# The arc of the parabola from to is rotated about the -axis. Find the area of the resulting surface. Show all steps in solving the appropriate integral.

This is how the answer was supposed to display.  I appologize for the delay.

There is insufficent information to solve this problem as the equation of the parabola, the portion of the arc that is rotated, and which axis it is rotated around is mission.

That said, I will provide an example of how to solve a problem such as this with data that I will make up.  So, to finish your question:

The arc of the parabola `y(x)=+-sqrt(x)` from x=1 to x=5 is rotated about the x-axis.  Find the area of the resulting surface.

In order to solve for area:

A = `int_a^b 2piy(x)sqrt(1+y'(x)^2)dx`

Take the derivative of the positive portion of our parabola (because the portion of the arc that we are interested in is for positive values of y):

`y'(x)=1/(2sqrt(x))`

Therefore:

`A=int_1^5 2pisqrt(x)sqrt(1+(1/(2sqrt(x)))^2)dx`

`A=piint_1^5sqrt((2sqrt(x))^2)sqrt(1+(1/4x))dx`

`A=piint_1^5sqrt(4x+1)dx`

Using integration by substitution:

z = 4x + 1, dz = 4 dx therefore dx = dz/4

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Hmm.. that did not display as it did for me before I pressed submit.  I will have to return later to fill in the areas that seem to have been deleted.

Approved by eNotes Editorial Team

There is insufficent information to solve this problem as the equation of the parabola, the portion of the arc that is rotated, and which axis it is rotated around is mission.

That said, I will provide an example of how to solve a problem such as this with data that I will make up.  So, to finish your question:

The arc of the parabola ``  from x=1 to x=5 is rotated about the x-axis.  Find the area of the resulting surface.

In order to solve for area:

` `

`` , a=1 and b=5

``

Using integration by substitution:

z = 4x + 1, dz = 4 dx therefore dx = dz/4

`A = pi/4 int_1^5 sqrt(z)dz`
` ` `A=pi/4 [(2/3)z^(3/2)]=pi/6[(4x+1)^(3/2)]_1^5`
`A=pi/6 ((4(5)+1)^(3/2)-(4(1)+1)^(3/2))=5.5`