Question

This is supposed to be able to be solved using double integrals, but I keep finding ways that don't work. Please Help!

Let S be the portion of the hemisphere defined by f(x,y)=square root(4-x^2-y^2) that lies above the plane z=1 but below the plane z=square root(3). Find the surface area of S.

Should I try converting to polar? and I think that one set of bounds should be from z=1 to z=sqrt(3), but then what would the other bounds be?

Answer 1

may i ask how you tried

Answer 2

I think definitely polar way is best

Answer 3

like just me give your setup
you don't have to work out fully

Answer 4

When I was thinking polar, my only problem was that then I would have three bounds, dz,dr, and dtheta
I also tried just using dz and dx, but got lost as to what my bounds would be

Answer 5

no i still think it is just dr and d theta

Answer 6

what if I use dz and dtheta, since the radius is always just the equation seeing as it is surface area

Answer 7

i think for dr it would be from 0 to 2
d theta would be from 0 to 2 pi

i'm not sure what we would do with z=1 to z=sqrt(3) think yet
That is interesting...

Answer 8

Where do you get the 2 from for dr?

Answer 9

we have an upper sphere
x^2+y^2+z^2=4

Answer 10

x^2+y^2+z^2=2^2

Answer 11

Yes, but when you input the values for z, you get a radius of 1 for the upper part and a radius of root 3 for the lower part

Answer 12

so maybe r is from 1 to sqrt(3)

Answer 13

and we still have theta from 0 to 2 pi

Answer 14

my only problem is that we are supposed to be finding surface area, and a radius that varies would create a filled solid

Answer 15

do you happen to have an answer to this problem?

Answer 16

All I have is an answer that someone else came up with, but I don't know if it is right. they got 2pi (sqrt(3)-⅓). This is using conversion to polar, but I don't remember exactly what they did

Answer 17

ok say we have \[x^2+y^2+z^2=4 => r^2+z^2=4 \]
so
if z is between 1 sqrt(3)
then r is also between 1 and sqrt(3)

since
\[z=\sqrt{4-r^2} \]
or let's write for r since we want to know what r equals when z equals those values above
\[r=\sqrt{4-z^2} \text{ so for } z=1 \text{ we have } r=\sqrt{3} \text{ and } \text{ when we have} \]
\[z=\sqrt{3} \text{ then we } r=1 \]

Answer 18

so do we use r instead of z?

Answer 19

i sorta think all it is
just doing \[\int\limits_{0}^{2 \pi} \int\limits_{1}^{\sqrt{3}} \frac{r}{\sqrt{4-r^2}} dr d \theta \]

Answer 20

it doesn't give the solution your friend got
but it i got something close
2pi(sqrt(3)-1)

Answer 21

why would the square root be on the bottom?

Answer 22

\[z_x= \frac{-x}{\sqrt{4-x^2-y^2}} \text{ and } z_y=\frac{-y}{\sqrt{4-x^2-y^2}}\]

Answer 23

I have to go to a meeting, but will be back in about an hour and a half. Thank you so much for helping me with this!

Answer 24

@satellite73 I hate double integrals.
Is my setup right?

Answer 25

I totally understand that feeling!
Yes the set up is right, at least I think so

Answer 26

Hey the more I think about it, the more I like my setup. Here is a similar problem.
http://math.stackexchange.com/questions/475801/find-the-surface-area-of-the-part-of-the-paraboloid-z-5-x2-y2-that-lies

Answer 27

do you understand how i got the radical part on bottom?

Answer 28

do you see they had z between 0 and 1
and that one guy used the relationship between z and r to figure out his r limits

Answer 29

I understand the radical on the bottom.
That is an awesome example, thank you for finding that! How did he find the dr/dx and the dr/dy though? I don't quite understand where he was getting that from.

Answer 30

do you mean dz/dx and dz/dy?

Answer 31

on i see the part you are talking about

Answer 32

yeah, I don't quite understand where that is coming from, or how he gets root(4r^2+1) out of it.

Answer 33

\[\sqrt{f_x+f_y+1}=\sqrt{((5-(x^2+y^2))_x)^2+((5-(x^2+y^2))_y)^2+1}\]
\[=\sqrt{(-2x)^2+(-2y)^2+1}\]

Answer 34

you can just use the that formula

Answer 35

and he replaced x^2+y^2 with r^2 to convert it over to polar form

Answer 36

oh, that makes sense!

Answer 37

Now I am getting \[\int\limits_{0}^{2\pi}\int\limits_{1}^{\sqrt{3}}r^2/\sqrt{4-r^2}dr d theta \]

Answer 38

\[\sqrt{(f_x)^2+(f_y)^2+1}=\sqrt{(\frac{-x}{\sqrt{4-(x^2+y^2)}})^2+(\frac{-y}{\sqrt{4-(x^2+y^2)}})^2+1}\]
\[\sqrt{\frac{x^2}{4-(x^2+y^2)}+\frac{y^2}{4-(x^2+y^2)}+\frac{4-(x^2+y^2)}{4-(x^2+y)^2}}\]
\[\sqrt{\frac{(x^2+y^2)+4-(x^2+y^2)}{4-(x^2+y^2)}}=\sqrt{\frac{4}{4-(x^2+y^2)}}=2 \frac{1}{\sqrt{4-(x^2+y^2)}}\]

Answer 39

how did you get r^2