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I have seen similar questions be asked but I'm uncertain if the same principles work for a horn torus. The question I have to ask has to do with Part (iii) of the following question. For Part (i) I stated that the expression is valid for $0 ≤ \theta ≤ 2\pi$ (since I don't think you could create the full circle otherwise, I think the range of θ has to be $2\pi$ long). For Part (ii) I rearranged the equation to get the equation of a circle with centre (0, 0.5) and radius of 0.5. For Part (iii), I said that the surface area A, of the torus produced is: $$A = \int_{0}^{ 2\pi}2\pi y*ds$$

I believe that is what my teacher was saying in his lecture notes but I am unsure if this integral calculates the surface area of the torus or the cross-sectional area of the original circle. I think that it is the former.

Through my working out, I end up with this surface area and this volume. I thought the volume could be calculated by integrating the surface area with respect to the radius of rotation $R$ (I'm sure a similar principle is applied when finding the volume of a sphere from the surface area of said sphere, or the area of a circle from the perimeter of said circle).

My only doubt is that I have seen equations that state the surface area of the torus as being equal to $(2\pi r)(2\pi R)$, and in the question both R and r should be equal to each other, where $r = 0.5$, which would make my answer for the surface area incorrect. I am unsure where I have gone wrong or if there is something I am not properly understanding.

Any help would be much appreciated.

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Draw a figure, and you will see that the interval $0\leq\theta\leq\pi$ produces the full circle $x^1+\bigl(y-{1\over2}\bigl)^2={1\over4}$. Apart from that $\theta$-values with $-\pi<\theta<0$ are not acceptable, as they would lead to $r<0$.

From $r(\theta)=\sin\theta$ it follows that $$x(\theta)=\sin\theta\cos\theta, \qquad y(\theta)=\sin^2\theta\ ,$$ which then leads to $x'^2(\theta)+y'^2(\theta)\equiv1$, hence $ds=d\theta$. Your formula for the area element is correct in the present situation. We therefore obtain $${\rm area}(S)=\int_0^\pi 2\pi y(\theta)\>d\theta=2\pi\cdot{\pi\over2}=\pi^2\ .$$

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