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Solid of revolution

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Could someone please explain how to do this question.

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.
 

Answers and Replies

AKG
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Option 1: compute the area bounded by those curves, then find the x co-ordinate of its centroid and find the circumference of the circle whose radius is the distance from this centroid to the line x=2. Multiply the area by the circumference to get your volume (Pappus's Second Theorem).

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)4 are rotated about the x-axis, and use this to solve the problem as you already know how to.
 
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AKG said:
Notice why this problem is the same as [rotating the lines] about the x-axis
Do you mean about the y-axis? If the original problem says to rotate the solid about the line x=2.
 
AKG
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Yes, you're right, the y-axis (the line x = 0).
 
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Thanks for your help :smile:
 
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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.
 
Last edited:
TD
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By using [itex]y = \sqrt {x+2}[/itex] and [itex]y = (x+2)^4[/itex], you 'moved' the functions two places to the left. Your bounds have to move then as well, so you integrate from y = -2 to y = -1.

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]
 
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Okay, the bounds move, but that results in an undefined answer, if my expansion above is correct.

Shouldn't it be y= 2,1. You can't get a negative answer from those equations.
 
TD
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Your expansion isn't really clear to me...
 
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I'm not sure how I can make it clearer, but when you rearrange [itex]y = \sqrt {x+2}[/itex], you get [itex]x^2 = y^4-4y^2+4[/itex].

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.
 
TD
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If you have 2 functions g(x) and h(x) and you want to find the solid of revolution of their difference (f(x) = h(x) - g(x)), the function f(x)² in the previous mentioned formula isn't h(x)²-g(x)² but (h(x) - g(x))² which isn't the same.
 
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okay, [itex]y = \sqrt {x+2}[/itex], is rearranged to [itex]x = y^2-2[/itex].

[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.
 
TD
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I'm sorry, I think I confused you and my last answer wasn't entirely correct.

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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