Evaluate integral for surface of a paraboloid

• ashina14
In summary, the integral given in the conversation is wrong and needs to be corrected using the formula for dS. The surface, z = x^2 + y^2, can be written in terms of parameters, u and v, and the correct integral in polar coordinates is given as \int_{r= 0}^\sqrt{2}\int_{\theta= 0}^{2\pi} |r^4cos(\theta)sin(\theta)|\sqrt{4r^2+ 1}r drd\theta, which evaluates to (1/420) (125√5 - 1).
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

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

ashina14 said:

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.

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.

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.

It's given that z = 1 though

ashina14 said:
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?

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!

1. What is a paraboloid?

A paraboloid is a three-dimensional shape that resembles a parabola in cross-section. It is formed by rotating a parabola around its axis.

2. What is an integral for the surface of a paraboloid?

An integral for the surface of a paraboloid is a mathematical calculation used to find the area of the curved surface of a paraboloid. It involves finding the sum of infinitely small rectangles that make up the surface.

3. How do you evaluate an integral for the surface of a paraboloid?

To evaluate an integral for the surface of a paraboloid, you can use a specific formula or integrate the function representing the paraboloid over the desired bounds. This can be done using methods such as integration by parts or substitution.

4. What is the importance of evaluating an integral for the surface of a paraboloid?

Evaluating an integral for the surface of a paraboloid can be useful in various fields such as engineering, physics, and mathematics. It allows for the calculation of important quantities such as surface area, volume, and moments of inertia.

5. Are there any real-world applications of evaluating integrals for the surface of a paraboloid?

Yes, there are many real-world applications of evaluating integrals for the surface of a paraboloid. For example, it can be used to calculate the surface area of a satellite dish, the volume of a water tank, or the strength of a curved bridge.

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