- #1

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Find the volume of the solid formed when the area between y=√x and y=x^4 is rotated about the line x=2.

I know how to do this when it's rotated about the x and y axes, but I'm not sure how to do it with a different line.

Thank you.

- Thread starter ~angel~
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- #1

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Find the volume of the solid formed when the area between y=√x and y=x^4 is rotated about the line x=2.

I know how to do this when it's rotated about the x and y axes, but I'm not sure how to do it with a different line.

Thank you.

- #2

AKG

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Option 2: Notice why this problem is the same as the problem of finding the volume of the solid formed when the area between y = √(x + 2) and y = (x + 2)

- #3

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Do you mean about the y-axis? If the original problem says to rotate the solid about the line x=2.AKG said:Notice why this problem is the same as [rotating the lines] about the x-axis

- #4

AKG

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Yes, you're right, the y-axis (the line x = 0).

- #5

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Thanks for your help

- #6

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I can't seem to get the answer. I'm using the 2nd method. I'm meant to be rotating about the y-axis, aren't I?

In that case,

x^2=(y^2-2)^2------------- x^2=((y^1/4)-2)^2

=y^4-4y^2+4 --------------= y-(4y1/4)+4

Then you integrate these? The bounds are y=0, y=1. But I end up with completely the wrong answer.

In that case,

x^2=(y^2-2)^2------------- x^2=((y^1/4)-2)^2

=y^4-4y^2+4 --------------= y-(4y1/4)+4

Then you integrate these? The bounds are y=0, y=1. But I end up with completely the wrong answer.

Last edited:

- #7

TD

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The volume of a solid of revolution, about the y-axis, is the given by:

[tex]\pi \int\limits_a^b {f\left( y \right)^2 dy} [/tex]

- #8

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Shouldn't it be y= 2,1. You can't get a negative answer from those equations.

- #9

TD

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Your expansion isn't really clear to me...

- #10

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When you rearrange [itex]y = (x+2)^4[/itex], you get [itex]x^2 = y-4y^(1/4)+4[/itex]

For the latter, that is the 4th root for the 2nd term.

- #11

TD

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- #12

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[itex]y = (x+2)^4[/itex] is rearranged to [itex]x = y^(1/4)-2[/itex]

Then, [itex][(y^2-2)-(y^(1/4)-2)]^2[/itex]

Then you get [itex]y^4-y^2+y^(1/16)[/itex]

But shouldn't you just integrate from the same bounds because your only changing the values of the x-axis.

- #13

TD

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Using f(x) = h(x) - g(x) doesn't change the area, but it does move it so the volume of the solid of revolution doesn't stay the same. Therefore, it would be indeed better to calculate the entire V of the solid of revolution of h(x) and then substract the one of g(x).

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