Question

Find the area of the surface S.

S is the part of hte paraboloid z = x^2 +y^2 cut off by the plane z = 1

Answer 1

I can do this using a rotation, but I'm not sure about the correct process with iterated integrals.

Answer 2

will the formula help?
\[\int\limits_{?}^{?}\int\limits_{?}^{?} _r \sqrt{[fx(x,y)]^2 +[fy(x,y)]^2 +1 } dA\]

Answer 3

That would help quite a bit actually. Although it looks like timo has got it now.

Answer 4

i found similar in my book i'll attach it very very similar

Answer 5

just use your numbers i guess

Answer 6

The only difference between this problem and that problem is the limits of integration. This one should go from 0 to 1 instead of 0 to 3 for dr.

Answer 7

Right that is why i said use your numbers to the OP :) and convert to polar

Answer 8

It's just the book doesn't explain the steps very well, so I thought I'd make the difference(s) explicit.

Answer 9

i don't understand how to find the limits.. like if its dydx aren't i supposed to put the y boundary in terms of y? like y=x^2 to y = x^3 or whatever? but i don't has that..

Answer 10

That's why we convert to polar where \(r^2=x^2+y^2\) and integrate from 0 to \(r\) since \(r\) is the radius of your paraboloid.

Answer 11

yup to king george convert to polar

Answer 12

so the equation would be r^2, but how do i find the new limit with polars? :(

Answer 13

Since we're in polar, we want to integrate from 0 to the maximum radius with respect to r, and from 0 to 2pi with respect to theta since it's over a circle.

Answer 14

It just so happens that the maximum radius is 1 in your problem, so we integrate from 0 to 1 in this particular problem.

Answer 15

here is a review on polar

also x= r cos theta and y = r sin theta

Answer 16

how do you know max radius is 1? :o

Answer 17

I've got to go now. I'm sure timo will be able to answer any more questions you have.

Good luck.

Answer 18

max radius is given cuzz z=1

Answer 19

thanks george!

Answer 20

ohhh okay

Answer 21

in the book example z=9 but yours is 1 If you take a slice of the parabaloid z=x^2+y^2 when z= 9 you get a circle of radious 3

In your case your z=1 so you get circle of radious 1 :)

Answer 22

\[\sqrt{4} \int\limits_{0}^{2\Pi}\int\limits_{0}^{1} \sqrt{r^2+1}r dr d \theta\]

Answer 23

is that right?

Answer 24

I got this
(5/6)*sqrt(5)*Pi-(1/6)*Pi

Answer 25

int(int(sqrt(1+4*r^2)*r, r = 0 .. 1), theta = 0 .. 2*pi)
That is how i set it up That is maple code btw

Answer 26

screen shot of how i set it up

Answer 27

intint(1+4x2+4y2)^.5dxdy

Answer 28

fx=2x , fy=2y
so in polar form int int (1+4r2)^.5rdrdtet=2pi*(1+4r2)1.5 /12=pi/6(5^1.5-1)

Answer 29

can you please explain how to integrate \[\int\limits_{0}^{1}\sqrt{1+4r^2}rdrd \theta\]

Answer 30

perhaps use a substitute of: tan(u) = 2r