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Problem solved
I'm not sure how easy this will be to understand without a diagram, but I don't know how to upload one :(
Let a and b be constants, with a > b > 0.A torus is formed by rotating the
circle (x - a)^2 + y^2 = b^2 about the y-axis.
The cross-section at y = h, where –b ≤ h ≤ b, is an annulus. The annulus has
inner radius x1 and outer radius x2 where x1 and x2 are the roots of
(x - a)^2 = b^2 + y^2(i) Find x1 and x2 in terms of h.
(ii) Find the area of the cross-section at height h, in terms of h.
(iii) Find the volume of the torus.
I think they're all up there.
I did part (i) and got x1 = a - \sqrt{b^2 - h^2}
and x2 = a + \sqrt{b^2 - h^2}, which is correct according to the answers.
I did part (ii) and got A = \pia\sqrt{b^2 - h^2}, which again is correct.
But for part 3 I thought V = \pia\int^{b}_{-b}\sqrt{b^2 - h^2}dh (sorry I don't know how to get rid of that itex in the integral)<br /> <br /> But in the answers it is V = \pia\int^{a}_{-a}\sqrt{b^2 - h^2}dh, and I am not sure what it&#039;s between -a and a. The answers are from a different source then the question so I&#039;m not sure if they have made a mistake or not.<br /> <br /> Any help will be much appreciated, thanks.
I'm not sure how easy this will be to understand without a diagram, but I don't know how to upload one :(
Homework Statement
Let a and b be constants, with a > b > 0.A torus is formed by rotating the
circle (x - a)^2 + y^2 = b^2 about the y-axis.
The cross-section at y = h, where –b ≤ h ≤ b, is an annulus. The annulus has
inner radius x1 and outer radius x2 where x1 and x2 are the roots of
(x - a)^2 = b^2 + y^2(i) Find x1 and x2 in terms of h.
(ii) Find the area of the cross-section at height h, in terms of h.
(iii) Find the volume of the torus.
Homework Equations
I think they're all up there.
The Attempt at a Solution
I did part (i) and got x1 = a - \sqrt{b^2 - h^2}
and x2 = a + \sqrt{b^2 - h^2}, which is correct according to the answers.
I did part (ii) and got A = \pia\sqrt{b^2 - h^2}, which again is correct.
But for part 3 I thought V = \pia\int^{b}_{-b}\sqrt{b^2 - h^2}dh (sorry I don't know how to get rid of that itex in the integral)<br /> <br /> But in the answers it is V = \pia\int^{a}_{-a}\sqrt{b^2 - h^2}dh, and I am not sure what it&#039;s between -a and a. The answers are from a different source then the question so I&#039;m not sure if they have made a mistake or not.<br /> <br /> Any help will be much appreciated, thanks.
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