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    I'm stuck with part 2.

    I keep getting the answer as 6 \pi^2 a.

    Basically, I assumed that the Area would be:

    A \ =  \ \iint\limits_S \mathbf{r}.\mathbf{dS} \ \mathrm{d}\theta\,\mathrm{d} \phi

    Eventually, I ended up with

    \mathbf{r}.\mathbf{dS} = (a+cos \phi)(a cos \phi + 1)\mathrm{d}\theta\,\mathrm{d} \phi

    Is my r.dS wrong, or am I going wrong somewhere else?

    From here all I had to do was integrate, and it was pretty straightforward to get to 6 \pi^2 a as an answer.
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    The surface area is just \displaystyle \int_S \text{d}S.
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    (Original post by wanderlust.xx)


    I'm stuck with part 2.

    I keep getting the answer as 6 \pi^2 a.

    Basically, I assumed that the Area would be:

    A \ =  \ \iint\limits_S \mathbf{r}.\mathbf{dS} \ \mathrm{d}\theta\,\mathrm{d} \phi

    Eventually, I ended up with

    \mathbf{r}.\mathbf{dS} = (a+cos \phi)(a cos \phi + 1)\mathrm{d}\theta\,\mathrm{d} \phi

    Is my r.dS wrong, or am I going wrong somewhere else?

    From here all I had to do was integrate, and it was pretty straightforward to get to 6 \pi^2 a as an answer.
    Your calculation is right but r.dS gives the suface integral and not
    the surface itself.
    As it is written calculate the dS=|\vec{dS}|=\sqrt{s_x^2+s_y^2+  s_z^2}
    then the surface T=\iint_S\ dS
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    So, wait, what's the difference between the surface area and the surface integral? I thought they were just the same? =/
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    (Original post by wanderlust.xx)
    So, wait, what's the difference between the surface area and the surface integral? I thought they were just the same? =/
    A surface is an integral over the surface, but the integrand can be any function you like, and scalar-valued or vector-valued. You can think of the surface integral as calculating an area over the surface, but attaching a weight to every point of the surface, with this weight being our function. But when finding the surface area, we naturally want this weight to be 1 everywhere.
 
 
 
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