Area of a parametric surface. (Multiple integral)

[maCrobo]

Homework Statement

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.

Homework Equations

\int_{}^{}{\int_{}^{}{\left| \left| \frac{\partial r}{\partial x}\times \; \frac{\partial r}{\partial y} \right| \right|\partial x\partial y}}

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.

Thanks in advance! :D

 

[ehild]

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.

Thanks in advance! :D

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

ehild

 

[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.

 

[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.

 

[ehild]

The boundaries look correct. What are your integration results?

ehild

 

[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?

 

[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

 

[maCrobo]

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

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.

 

[ehild]

I chose z=√xy

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

 

[maCrobo]

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

Ops...

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

The answer should be 4...

 

[ehild]

√2(√2³+√2)≠10

ehild

 

[maCrobo]

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

 

[ehild]

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

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

ehild