Find the surface area S

**downthesun01** · December 8th 2010, 11:30 PM

S is the upper cap cut from the sphere $x^2+y^2+z^2=25$ by the cylinder $x^2+y^2=16$

Ok, so

$\displaystyle \iint_{R} \frac{|\nabla f|}{|\nabla f * p|}dA$

$R: x^2+y^2\leq 16$ on the xy-plane[/tex]

P is a vector normal to R, so p=k

$\nabla f=2xi+2yj+2zk$

$|\nabla f|=2\sqrt{2}$

$|\nabla f * p|=2z$

So, now I have

$\displaystyle \iint_{R} \frac{2\sqrt{2}}{2z}dA$

After factoring out constants and rewriting z as a function of x and y I have

$\displaystyle \sqrt{2}\iint_{R}\frac{1}{\sqrt{25-x^2-y^2}}dA$

Now, converting to polar coordinates, I get

$\displaystyle \sqrt{2}\int_{0}^{2\pi}\int^{4}_{0}\frac{1}{\sqrt{ 25-r^2}}rdrd\theta$

Ok, I let $u=25-r^2$ . Then, $du=-2rdr$

Which then gives me

$\displaystyle \sqrt{2}\int_{0}^{2\pi}\frac{1}{2}\int^{4}_{0}\fra c{-1}{\sqrt{u}}dud\theta$

Then I get

$\displaystyle \sqrt{2}\int_{0}^{2\pi}[ -\sqrt{25-r^2}}]^{4}_{0} d\theta$

Now, I have

$\displaystyle \sqrt{2}\int_{0}^{2\pi}2d\theta$

So

$\frac{\sqrt{2}}{2}[\theta]^{2\pi}_{0}$

$=\frac{\sqrt{2}}{2}(2\pi)$

The answer in the book is $20\pi$ . can someone help me find my mistake(s)? Thanks.

**HallsofIvy** · December 9th 2010, 02:26 AM

Originally Posted by downthesun01

S is the upper cap cut from the sphere $x^2+y^2+z^2=25$ by the cylinder $x^2+y^2=16$

Ok, so

$\displaystyle \iint_{R} \frac{|\nabla f|}{|\nabla f * p|}dA$

$R: x^2+y^2\leq 16$ on the xy-plane[/tex]

P is a vector normal to R, so p=k

$\nabla f=2xi+2yj+2zk$

$|\nabla f|=2\sqrt{2}$

Well, here's a problem: $|2xi+ 2yj+ 2zk|= \sqrt{4x^2+ 4y^2+ 4z^2}= 2\sqrt{x^2+ y^2+ z^2}= 2\sqrt{25}= 10$ , not " $2\sqrt{2}$ ".

$|\nabla f * p|=2z$

So, now I have

$\displaystyle \iint_{R} \frac{2\sqrt{2}}{2z}dA$

After factoring out constants and rewriting z as a function of x and y I have

$\displaystyle \sqrt{2}\iint_{R}\frac{1}{\sqrt{25-x^2-y^2}}dA$

Now, converting to polar coordinates, I get

$\displaystyle \sqrt{2}\int_{0}^{2\pi}\int^{4}_{0}\frac{1}{\sqrt{ 25-r^2}}rdrd\theta$

Ok, I let $u=25-r^2$ . Then, $du=-2rdr$

Which then gives me

$\displaystyle \sqrt{2}\int_{0}^{2\pi}\frac{1}{2}\int^{4}_{0}\fra c{-1}{\sqrt{u}}dud\theta$

Then I get

$\displaystyle \sqrt{2}\int_{0}^{2\pi}[ -\sqrt{25-r^2}}]^{4}_{0} d\theta$

Now, I have

$\displaystyle \sqrt{2}\int_{0}^{2\pi}2d\theta$

So

$\frac{\sqrt{2}}{2}[\theta]^{2\pi}_{0}$

$=\frac{\sqrt{2}}{2}(2\pi)$

The answer in the book is $20\pi$ . can someone help me find my mistake(s)? Thanks.

**downthesun01** · December 9th 2010, 02:30 AM

Thanks. I'll try reworking it now

**moderata** · December 13th 2010, 08:47 PM

when you are using your $u substitution$ , note:

$1/sqrt(u) = u^-^1^/^2$ so when you are taking it's anti-derivative, you will gain a 2 as well. It should look like this: $2u^1^/^2$ . After taking the derivative of this value you will be back to your original answer. This will also be what gets the $5/2$ back to being a whole 5, which then allows it to become a 10 again, resulting in your answer of $20pi$ as per the text.

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Find the surface area S

Also, check your u substitution

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