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The question asks to integrate the vector field $$ F(x,y,z)={1\over\sqrt{x^2+y^2}}(y,-y,1) $$ over the paraboloid $$z=1-x^2-y^2.$$ over the area defined by $0\leq z\leq1$ which is $x^2+y^2\leq1$. Using $X=(x,y,1-x^2-y^2)$ as the surface I obtained $$ \begin{align} {\partial X\over\partial x}&=(1,0,-2x)\\ {\partial X\over\partial y}&=(0,1,-2y)\\ N={\partial X\over\partial x}\times {\partial X\over\partial y}&=(2x,2y,1)\\ \end{align} $$ and the integral of $F$ is then $$ \iint_R F(x,y,z)\cdot N dx\;dy = \int_0^{2\pi}\int_0^1 {2xy-2y^2+1\over r} r\;dr\;d\theta\\ =\frac13\int_0^{2\pi}(2\sin\theta\cos\theta-2\sin^2\theta+1) \;d\theta\\ =\frac13\int_0^{2\pi}(\sin2\theta+\cos2\theta)\;d\theta=0.\\ $$ However, the book gives the answer $4\pi/3$.

I've been over the calculations repeatedly without avail. Can anyone point out my error?

This is exercise 12 of XII ยง3 of "Calculus of Several Variables" by Serge Lang, third edition, on page 341.

Suzu Hirose
  • 11,660

1 Answers1

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The error is here:

$$\int_0^{2\pi}\int_0^1{2xy-2y^2+1\over r} r\;dr\;d\theta\\=\frac13\int_0^{2\pi}(2\sin\theta\cos\theta-2\sin^2\theta+1) \;d\theta\\$$ I performed the $r^2$ integral for the $xy$ and $y^2$ terms correctly, but the final $1$ on top of that integrand has no $r^2$ in it so it should have been integrated as $\int_0^1 1 dr=1$ not $\frac13$ as I've written above.

This gives the extra $4\pi/3$ as required.

(As has happened to me several times now, I found the error after typing this question in. I'll leave it as a spoiler for people who want to see if they can spot the error as a puzzle, although it jumped right out at me after I'd typed all of the question and then read it back.

I've done this before, prepared a question here and then spotted my mistake. To now I have discarded the draft question, but I thought this time I would ask the question and then self-answer it, in case it's some use for others.)

Suzu Hirose
  • 11,660