Integral

[bugatti79]

Hi Folks,

Need to evaluate the surface integral

\(\displaystyle \displaystyle \int \int_\sigma f(x,y,z) dS\) for \(\displaystyle f(x,y,z)=z+1\) where the surface os the upper hemisphere \(\displaystyle z=\sqrt{1-x^2-y^2}\)

I realise this can be done by parameterising the surface using θ and ∅. However, is it possible to use this other method
\(\displaystyle \displaystyle \int \int_{\sigma} f(x,y,z) dS=\int \int_R f(x,y,g(x,y))\sqrt{z_x^2+z_y^2+1}dA \) (1)

I calculate the RHS of the above equation to be

\(\displaystyle \displaystyle \int \int_R (1+ \frac{1}{\sqrt{1-x^2-y^2}})dA \)

Is this correct so far?

Thanks

 

[galactus]

I think polar may be the way to go. Usually, anytime you see \(\displaystyle x^{2}+y^{2}\), polar can simplify things a bit.

Yes, it would appear you have it thus far.

R is the circular region enclosed by \(\displaystyle x^{2}+y^{2}=1\)

\(\displaystyle \displaystyle \int\int(z+1)dS=\int\int (\sqrt{1-x^{2}-y^{2}}+1)\)\(\displaystyle \sqrt{\frac{x^{2}}{1-x^{2}-y^{2}}+\frac{y^{2}}{1-x^{2}-y^{2}}+1}dA\)

\(\displaystyle =\displaystyle \int\int\frac{\sqrt{1-x^{2}-y^{2}}+1}{\sqrt{1-x^{2}-y^{2}}}dA\)

Continuing where you left off:

\(\displaystyle =\displaystyle \lim_{a\to 1^{-}}\int_{0}^{2\pi}\int_{0}^{a}\left(1+\frac{1}{\sqrt{1-r^{2}}}\right)rdrd\theta\)

\(\displaystyle =\displaystyle \lim_{a\to 1^{-}}2\pi\left(\frac{1}{2}a^{2}-\sqrt{1-a^{2}}+1\right)=\)

Now, take the limit and you have it.

 

[bugatti79]

I think polar may be the way to go. Usually, anytime you see \(\displaystyle x^{2}+y^{2}\), polar can simplify things a bit.

Yes, it would appear you have it thus far.

R is the circular region enclosed by \(\displaystyle x^{2}+y^{2}=1\)

\(\displaystyle \displaystyle \int\int(z+1)dS=\int\int (\sqrt{1-x^{2}-y^{2}}+1)\)\(\displaystyle \sqrt{\frac{x^{2}}{1-x^{2}-y^{2}}+\frac{y^{2}}{1-x^{2}-y^{2}}+1}dA\)

\(\displaystyle =\displaystyle \int\int\frac{\sqrt{1-x^{2}-y^{2}}+1}{\sqrt{1-x^{2}-y^{2}}}dA\)

Continuing where you left off:

\(\displaystyle =\displaystyle \lim_{a\to 1^{-}}\int_{0}^{2\pi}\int_{0}^{a}\left(1+\frac{1}{\sqrt{1-r^{2}}}\right)rdrd\theta\)

\(\displaystyle =\displaystyle \lim_{a\to 1^{-}}2\pi\left(\frac{1}{2}a^{2}-\sqrt{1-a^{2}}+1\right)=\)

Now, take the limit and you have it.

Thanks, I think there is a 1 that shouldnt be there as highlighted in red..? I calculate the answer to be 3*pi...?

 

[Deleted member 4993]

No - that 1 is correct. It comes from the limits of definite integral (r = 0).

 

[bugatti79]

No - that 1 is correct. It comes from the limits of definite integral (r = 0).

\(\displaystyle =\displaystyle \lim_{a\to 1^{-}}\int_{0}^{2\pi}\int_{0}^{a}\left(1+\frac{1}{\sqrt{1-r^{2}}}\right)rdrd\theta\)

My approach to this inner integral is where a=1


\(\displaystyle =\displaystyle \int_{0}^{1}\left(r+\frac{r}{\sqrt{1-r^{2}}}\right)dr\) using substitution for the second term by letting u=1-r^2 and so on. I dont see where the 1 comes from?

thanks

 

[Deleted member 4993]

\(\displaystyle =\displaystyle \int_{0}^{a}\left(\frac{r}{\sqrt{1-r^{2}}}\right)dr\)

\(\displaystyle = \ - \ \left [ \sqrt{1-r^2}\ \right ]_0^a \)

what do you get from here?

 

[bugatti79]

\(\displaystyle =\displaystyle \lim_{a\to 1^{-}}\int_{0}^{2\pi}\int_{0}^{a}\left(1+\frac{1}{\sqrt{1-r^{2}}}\right)rdrd\theta\)

My approach to this inner integral is where a=1


\(\displaystyle =\displaystyle \int_{0}^{1}\left(r+\frac{r}{\sqrt{1-r^{2}}}\right)dr\) using substitution for the second term by letting u=1-r^2 and so on. I dont see where the 1 comes from?

thanks

I get \(\displaystyle \displaystyle \int^{2 \pi}_{0} [\frac{r^2}{2}-\sqrt{1-r^2}]^{a}_{0} d\theta=\int^{2 \pi}_{0} [ (1/2-0)-(0-1)] d\theta=3 \pi\)

where a =1......?

 

[bugatti79]

I get \(\displaystyle \displaystyle \int^{2 \pi}_{0} [\frac{r^2}{2}-\sqrt{1-r^2}]^{a}_{0} d\theta=\int^{2 \pi}_{0} [ (1/2-0)-(0-1)] d\theta=3 \pi\)

where a =1......?

Any one willing to check?

Thanks