Evaluate integral for surface of a paraboloid

[ashina14]

Homework Statement

Evaluate s∫∫ lxyzl dS, where S is part of the surface of paraboloid
z = x2 + y2, lying below the plane z = 1

Homework Equations

The Attempt at a Solution

since z=1 and x2+y2=z,

therefore integral becomes 0∫^1 0∫^(1-x2) xyz dy dx

which solves to 1/8. Apprently this is wrong but I don't know how

 

[LCKurtz]

Homework Statement

Evaluate s∫∫ lxyzl dS, where S is part of the surface of paraboloid
z = x2 + y2, lying below the plane z = 1

Use the X² icon above the advanced editing box for superscripts: z = x² + y².

Homework Equations

The Attempt at a Solution

since z=1 and x2+y2=z,

therefore integral becomes 0∫^1 0∫^(1-x2) xyz dy dx

which solves to 1/8. Apprently this is wrong but I don't know how

You could start by looking up the formula for dS and surface integrals in your text.

 

[HallsofIvy]

The surface, z = x^2 + y^2, can be written in terms of parameters,u, v, as
x= u, y= v, z= u^2+ v^2 so that any point on the surface can be written in terms of the position vector \vec{r(u, v)}= u\vec{ix}+ v\vec{j}+ (u^2+ u^2)\vec{k}.

The derivatives with respect to the parameters are
\vec{r_u}= \vec{i}+ 2u\vec{j} and \vec{r_v}= \vec{j}+ 2v\vec{k}
Those are two vectors tangent to the surface, and their cross product,
\vec{r_v}\times\vec{r_u}= 2u\vec{i}+ 2v\vec{j}+ \vec{k}
times dudv, is the "vector differential of surface area". Its length, dS= \sqrt{4u^2+ 4v^2+ 1}dudv is the "differential of surface area".

So \int |xyz|dS= \int\int |uv(u^2+ v^2)(\sqrt{4u^2+ 4v^2+ 1}dudv
Seeing all those sum of squares I would be inclined to put it in "polar coordinates with u= rcos(\theta), v= r sin(\theta). z goes from 0 up to z= u^2+ v^2= r^2= 2 so that becomes
\int_{r= 0}^\sqrt{2}\int_{\theta= 0}^{2\pi} |r^4cos(\theta)sin(\theta)|\sqrt{4r^2+ 1}r drd\theta

To handle the absolute value, you will need to determine where cos(\theta)sin(\theta) is positive and where negative.

 

[ashina14]

It's given that z = 1 though

 

[LCKurtz]

It's given that z = 1 though

What kind of response is that? To whom are you responding?

If you are responding to my post, I asked you if you knew the formula for dS because your integral is completely wrong. And you haven't answered.

And I'm guessing you don't understand HallsofIvy's reply because you may not have had general parameterizations yet. But until you give us a real reply, and quote to whom you are replying, how are we to know?

 

[ashina14]

Sorry LCKurtz, I forgot to mention, it is a reply to Hallsofly, as they've assumed z=2 whereas the question states it is 1. In response to your question, I completely understand where I went wrong now, looked up the formula, followed Hallsofly's method and got the answer of (1/420) (125√5 - 1). Thanks a lot guys!