# Area of a parametric surface. (Multiple integral)

1. Dec 31, 2011

### maCrobo

1. The problem statement, all variables and given/known data
First exercise:
Compute the surface area of that portion of the sphere x2+y2+z2=a2 lying within the cylinder x2+y2=ay, where a>0.
Second one:
Compute the area of that portion of the surface z2=2xy which lies above the first quadrant of the xy-plane and is cut off by the planes x=2 and y=1.

2. Relevant equations
$\int_{}^{}{\int_{}^{}{\left| \left| \frac{\partial r}{\partial x}\times \; \frac{\partial r}{\partial y} \right| \right|\partial x\partial y}}$

3. The attempt at a solution

First (exercise) parametrization: $r\left( x,y \right)=x\; i+y\; j+\sqrt{a^{2}-x^{2}-y^{2}} \; k$
By using the "relevant equation" I wrote before, I get the following integral: $\int_{}^{}{\int_{}^{}{\frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}}\partial x\partial y}}$

Second (exercise) parametrization: $r\left( x,y \right)=x\; i+y\; j+\sqrt{2xy} \; k$
By using again the "relevant equation" I wrote before, I get the following integral: $\int_{}^{}{\int_{}^{}{\frac{x+y}{\sqrt{2xy}} \partial x \partial y}}$

The boundaries of the integrals can be easily got from the text of the exercise, but I won't write them not to influence your reasoning. I think there are my mistakes.
Anyway, my results are wrong, so I please you to try and tell me your results so to understand where I got lost.

2. Dec 31, 2011

### ehild

We can see what you did wrong only when you show it. So what are the boundaries?

ehild

3. Dec 31, 2011

### maCrobo

Ok.
As for the first exercise: $\int_{0}^{1}{\left( \int_{0}^{2}{\frac{x+y}{\sqrt{2xy}}\partial x} \right) \partial y}$

In fact the projection of the surface on the xy-plane is a rectangle.

As for the second exercise: $\int_{0}^{a}{\left( \int_{-\sqrt{ay-y^{^{2}}}}^{\sqrt{ay-y^{2}}}{\frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}} \partial x} \right)\partial y}$

I found the second integral a bit hard in this shape, so I tried to turn the variables in polar coordinates and I got:

$\int_{0}^{\pi }{\left( \int_{0}^{\mbox{asin}\theta }{\frac{ar}{\sqrt{a^{2}-r^{2}}} \partial r} \right) \partial \theta }$
Where x=rcosθ and y=rsinθ. Then the boundary become [0,π] for θ and I evaluated that of r by plugging x(r,θ) and y(r,θ) in the cylinder equation, so that I got r=asinθ.

4. Jan 1, 2012

### DavidAlan

This is only a conventional point, but do not get in the habit of using partials in replace of "d" in integrals. That partial symbol is kind of sacred. It either refers to derivatives, Jacobian determinants, or the boundaries of open sets. When you use it in the sense of differentials, that be scary.

5. Jan 1, 2012

### ehild

The boundaries look correct. What are your integration results?

ehild

6. Jan 2, 2012

### maCrobo

Good news. I did the first one... the problem was the book that has missed a bracket in the answer.

As for the second exercise, the integral I wrote here was wrong. The right one, I checked it, is: $\int_{0}^{1}{\int_{0}^{2}{\sqrt{\frac{y^{2}+x^{2}4xy}{4xy}}}dxdy}$
I don't manage to come out with a result from this last integral. I even don't manage to get the primitive.
Any idea?

7. Jan 2, 2012

### ehild

For the function z^2=2xy, you have to integrate (x+y)/√(2xy), as you wrote in your original post. It was correct. Think: The function itself is symmetric to x<=>y. So should be dA.
And DavidAlan is right, use dx, dy in the integrals.

ehild

8. Jan 3, 2012

### maCrobo

I chose z=√xy so when I differentiate with respect to dx first and dy after, I get y/(2√xy) and x/(2√xy). Then I have the result of the cross product of the partial derivatives, which is -y/(2√xy) i -x/(2√xy) j +1k and finally the norm of this vector is $\int_{0}^{1}{\int_{0}^{2}{\sqrt{\frac{y^{2}+x^{2}+4 xy}{4xy}}}dxdy}$.
I don't come out with the one I wrote previously

P.S.: in the post #6 there is a + sign missing between the x^2 and the 4xy.

9. Jan 3, 2012

### ehild

See your original post: It was z2=2xy.

10. Jan 4, 2012

### maCrobo

Ops...

Anyway I tried again, this are my calculations:
http://desmond.imageshack.us/Himg694/scaled.php?server=694&filename=fotodel040112alle1055.jpg&res=medium [Broken]

Last edited by a moderator: May 5, 2017
11. Jan 4, 2012

### ehild

√2(√23+√2)≠10

ehild

12. Jan 4, 2012

### maCrobo

I got it... I'm an idiot...
Anyway: DONE!

13. Jan 4, 2012

### ehild

It was just a silly mistake, I do myself quite often. You did a god job.

ehild