Surface Integral finding the limits

[Spout]

Homework Statement

I need to evaulate ∫ ∫_S dS where S is the surface z = x² + y², 0 ≤ z ≤ 4.

Homework Equations

dS = √( 1 + ƒ²_x + ƒ²_y)dxdy

The Attempt at a Solution

dS = √( 1 + 4x² + 4y²)dxdy

here's the problem

what are the limits to the surface integral? no clue.. dx means i should find a limit for x but i see no obvious way to, also i could change this to cylindrical coordinates but still don't know what the limits would be.

∫ ∫ √( 1 + 4x² + 4y²)dxdy

 

[DryRun]

Hi Spout

The first thing you should do is draw the graph, which is a paraboloid.

Put z = 4 in the equation of the surface, S. You will get a circle with radius 2.

Since you are projecting onto the x-y plane, find $z_x$ and $z_y$ hence find the surface area:
$$\int \int \sqrt{(z_x)^2+(z_y)^2+1}\,.dxdy $$
Then, as the projection is a circle, it is easier to evaluate this double integral using polar coordinates.

 

[Spout]

This is quite embrassing but what does "Since you are projecting onto the x-y plane, find zx and zy hence find the surface area" mean

up to now i only know the process of solving these problems i don't know what partial derivative represents or means, so I'm unable to comprehend this sentence could you give a tiny bit of a hint :)?

 

[algebrat]

When doing these, one crucial step that is often unavoidable is visualizing the region. I've thought about some ways to manipulate limits of integration without visualizing, but I think it would make things much harder. So you should learn to visualize the regions the describe.

You have a bowl shape that is parabolic (paraboloid is the term in this dimension), that is, consider the y=0 plane, then z=x^2 is parabolic (the term for curves). The bowl occupies space between z=0...4. At z=0, x and y must be zero. At the z=4 plane, you have the circle x^2+y^2=4, that is, radius 2. You had the right integrand, √(1+4x^2+4y^2) dxdy.

So to find the region that paramet(e)rizes the bowl via x and y, we need to know which values of x and y will garauntee we hit all points of the bowl exactly once. So that is the projection of the bowl down (and/or up) onto the xy plane. The projection is of course the disk, x^2+y^2≤ 4. The limits would be hard to integrate with, so our look would be much better if we use a change of coordinates, from rectangular to polar coordinate.

So we want to map from (x,y) to (r,θ). Then the integrand would be √(1+r^2), while dxdy becomes rdrdθ. Why don't you try to find out what the limits of integration would be.

 

[Spout]

I see

well I am guessing 0 ≤ r ≤ 2 because this covers the min and max radius

and 0 ≤ θ ≤ 2pi because this covers all points angular-ly

then the integral would be

∫(0 to 2pi) ∫(0 to 2) [ √( 1 + 4r²cos²θ + 4r²sin²θ) ] rdrdθ

∫(0 to 2pi) ∫(0 to 2) [ √( 1 + 4r²) ] rdrdθ

:)? amirite

 

[Spout]

it turned out correct, you guys are legends