Physgeek64 said:Homework Statement
Find the area of the part of z^2=xy that lies inside the hemisphere x^2+y^2+z^2=1, z>0
Homework Equations
da= double integral sqrt(1+(dz/dx)^2+(dz/dy)^2))dxdy
The Attempt at a Solution
(dz/dx)^2=y/2x
(dz/dy)^2=x/2y
=> double integral (x+y)(sqrt(2xy)^-1/5) dxdy
Now I'm guessing that a change of coordinates will be useful here. I was thinking spherical coordinates, due to the presence of a sphere. But I'm not too sure?
many thanks :)
I thought I had? I don't see how I can take that integral any further?Ray Vickson said:You have made a good start. What is stopping you from going ahead? It is much too early to guess about what kind of coordinate transformation to make: write down the complete details of your final double integral first, then decide on any possible changes of variables.
Physgeek64 said:I thought I had? I don't see how I can take that integral any further?
Thank you so much for replying- I really appreciate it :)
Ohh of course. So would the limits be x=-sqrt(1-y^2) to x=sqrt(1-y^2) and y from -1 to 1?Ray Vickson said:But: you are only half-finished. One of the most important aspects (that you have not yet addressed) is: what is the integration region in the (x,y)-plane? Without that, you can go nowhere.
Besides: your expression for dA is incorrect: it is not ##(x+y)/\sqrt{2xy} \: dx \, dy##.
Physgeek64 said:Ohh of course. So would the limits be x=-sqrt(1-y^2) to x=sqrt(1-y^2) and y from -1 to 1?
Is it not?
I got dz/dx=y/sqrt(2xy) and dz/dy=x/sqrt(2xy)
so 1+(dz/dx)^2+(dz/dy)^2= 1+y/2x+x/2y = (2xy+y^2+x^2)/2xy = (x+y)^2/2xy
so sqrt(1+(dz/dx)^2+(dz/dy)^2) = (x+y)/sqrt(2xy) ?
I am known for making a lot of algebraic mistakes so its more than possible that I've don't this wrong, but I can't see where
Thank you :)
Ray Vickson said:Your region is still incorrect. Instead of writing down the first thing that pops into your mind, take the time to it out carefully and in detail from first principles. (1) What determines the integration region in (x,y) - space? (2) How would you express this algebraically? (3) You will get a boundary curve in (x,y)-space. In terms of its algebraic characterization, what type of curve will you get?
You may need to devote quite a bit of time and effort to this part of the problem, in which case there are no shortcuts. It is more important to get it right than to save some time.
I would bet that you have worked out such problems before, in other courses, or maybe earlier in the current course.
And no: you still do not have the correct expression for dA. One last hint: your derivatives ##\partial z/\partial x## and ##\partial z/\partial y## are wrong.
Ray Vickson said:Your region is still incorrect. Instead of writing down the first thing that pops into your mind, take the time to it out carefully and in detail from first principles. (1) What determines the integration region in (x,y) - space? (2) How would you express this algebraically? (3) You will get a boundary curve in (x,y)-space. In terms of its algebraic characterization, what type of curve will you get?
You may need to devote quite a bit of time and effort to this part of the problem, in which case there are no shortcuts. It is more important to get it right than to save some time.
I would bet that you have worked out such problems before, in other courses, or maybe earlier in the current course.
And no: you still do not have the correct expression for dA. One last hint: your derivatives ##\partial z/\partial x## and ##\partial z/\partial y## are wrong.
Ahh my mistake. It was meant to be z^2=2xy in the original post. Sorry!vela said:You wrote ##z^2 = xy## in the original post. Now you're saying ##z = \sqrt{2xy}##. Which is correct?
Physgeek64 said:Ahh my mistake. It was meant to be z^2=2xy in the original post. Sorry!
I'm really sorry :( I didn't mean to mistype it- I'm just not that good with computers.Ray Vickson said:That changes everything! It changes the (x,y)-region, it changes the integrand---everything.
Physgeek64 said:I'm really sorry :( I didn't mean to mistype it- I'm just not that good with computers.
But I'm lost as to where to go from here...
Ray Vickson said:All my previous advice still holds: (1) work out the (x,y)-region carefully, from first principles. (2) Double-check your x and y-derivatives of ##z = \sqrt{2} \sqrt{xy}##. (3) Write down in detail the integration involved in getting your final answer. (4) Then, and only then, worry about how to actually do the integral or integrals.
Beyond that, I am not permitted to say more, at least by my interpretation of the PF rules.
The equation for the area of z^2=xy inside a hemisphere is A = 2πr^2, where r is the radius of the hemisphere.
To calculate the surface integral for z^2=xy inside a hemisphere, you can use the formula ∫∫√(1+(∂z/∂x)^2+(∂z/∂y)^2)dA, where the limits of integration are determined by the shape and orientation of the hemisphere.
The surface integral for z^2=xy inside a hemisphere represents the total surface area of the hemisphere. This can have applications in various fields, such as engineering and physics, where calculating surface areas is important for understanding and solving problems.
No, the surface integral for z^2=xy inside a hemisphere cannot be used to find the volume of the hemisphere. To find the volume, you would need to use the formula V = (2/3)πr^3, where r is the radius of the hemisphere.
Yes, the surface integral for z^2=xy inside a hemisphere has applications in various fields such as electromagnetism, fluid mechanics, and heat transfer. It can also be used to calculate the surface area of other curved shapes, not just hemispheres.