How Do You Determine the Limits for Surface Integrals?

[galaxy_twirl]

Homework Statement

Homework Equations

∫∫_D F((r(u,v))⋅(r_u x r_v) dA

The Attempt at a Solution

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I got stuck after finding the above, at where the double integrals are. :(

May I know how do I find the limits of this? (I always have trouble finding the limits to sub into the integrals. ><)

Thank you! :)

 

[LCKurtz]

You have chosen u = x and v = y. This is just like a problem you posted earlier. Draw a picture of the slanted plane in the first octant and use the triangle in the xy plane for your limits.

[Edit, added] One thing that may be causing you confusion is the use of $u,v$ parameters when they are unnecessary. You usually use those when there is a reason to change coordinate systems. In a problem like this, $x$ and $y$ are natural parameters and there is no need to use $u$ and $v$. Just let $x=x,~y=y,~z= 2-2x-y$ and$$r(x,y)=\langle x,y,2-2x-y\rangle$$and look in the $xy$ plane for the limits.

 

[galaxy_twirl]

You have chosen u = x and v = y. This is just like a problem you posted earlier. Draw a picture of the slanted plane in the first octant and use the triangle in the xy plane for your limits.

[Edit, added] One thing that may be causing you confusion is the use of $u,v$ parameters when they are unnecessary. You usually use those when there is a reason to change coordinate systems. In a problem like this, $x$ and $y$ are natural parameters and there is no need to use $u$ and $v$. Just let $x=x,~y=y,~z= 2-2x-y$ and$$r(x,y)=\langle x,y,2-2x-y\rangle$$and look in the $xy$ plane for the limits.

Oh dear! Haha. I didn't realize that this question is similar to the one I have asked earlier. I am currently focusing on past year papers as my Maths exam is just 2 days away. Sigh~ I find that I always have a problem visualising the domain... because I didn't know the domain was a triangle until you told me above and after doing some look-up on Google. >< Sheesh. :(

I will take note of what you have said and try again. :) I will post back here again!

 

[STEMucator]

Oh dear! Haha. I didn't realize that this question is similar to the one I have asked earlier. I am currently focusing on past year papers as my Maths exam is just 2 days away. Sigh~ I find that I always have a problem visualising the domain... because I didn't know the domain was a triangle until you told me above and after doing some look-up on Google. >< Sheesh. :(

I will take note of what you have said and try again. :) I will post back here again!

A linear equation in three variables represents a plane.

The given equation $z = 2 - 2x - y$ is a plane. Another example would be $x + y - z = 1 \Rightarrow z = x + y - 1$.

If you want to have an easier time visualizing, then you can easily solve for the intersections. Consider the points $(x, 0, 0), (0, y, 0), (0, 0, z)$ for the plane equation in your problem.

Choosing the point $(x, 0, 0)$, you would get $z = 2 - 2x - y \Rightarrow x = 1$. In a similar fashion you would get $y = 2$ and $z = 2$. Plotting these solutions on a 3D graph and connecting the dots will give you a nice visual of the plane.

This will also help you visualize the projection of the plane onto the x-y plane. Drawing this projection in the x-y plane will give you your limits for integrals.

 

[galaxy_twirl]

You have chosen u = x and v = y. This is just like a problem you posted earlier. Draw a picture of the slanted plane in the first octant and use the triangle in the xy plane for your limits.

[Edit, added] One thing that may be causing you confusion is the use of $u,v$ parameters when they are unnecessary. You usually use those when there is a reason to change coordinate systems. In a problem like this, $x$ and $y$ are natural parameters and there is no need to use $u$ and $v$. Just let $x=x,~y=y,~z= 2-2x-y$ and$$r(x,y)=\langle x,y,2-2x-y\rangle$$and look in the $xy$ plane for the limits.

I see. Thank you. My teacher changes them to r(u,v) for the chapter of surface integrals, hence I would always change it to r(u,v), though I don't know why my teacher uses the u,v coordinate system. I think I should ask him some day. Haha.

Ah yes, I got the triangle in the xy-plane as shown below:

Just wondering, how can I be very sure that I have placed the limits on my ∫ ∫ correctly? My integral equation is below:

Thank you! :)

 

[galaxy_twirl]

A linear equation in three variables represents a plane.

The given equation $z = 2 - 2x - y$ is a plane. Another example would be $x + y - z = 1 \Rightarrow z = x + y - 1$.

If you want to have an easier time visualizing, then you can easily solve for the intersections. Consider the points $(x, 0, 0), (0, y, 0), (0, 0, z)$ for the plane equation in your problem.

Choosing the point $(x, 0, 0)$, you would get $z = 2 - 2x - y \Rightarrow x = 1$. In a similar fashion you would get $y = 2$ and $z = 2$. Plotting these solutions on a 3D graph and connecting the dots will give you a nice visual of the plane.

This will also help you visualize the projection of the plane onto the x-y plane. Drawing this projection in the x-y plane will give you your limits for integrals.

Thanks, Zondrina! Your method is really effective! :DI think I can now at least find the intersection points of the lines on the x-y-z axes, though I feel it can be quite difficult to plot/sketch/draw a 3D figure on a 2D piece of paper.

 

[LCKurtz]

I see. Thank you. My teacher changes them to r(u,v) for the chapter of surface integrals, hence I would always change it to r(u,v), though I don't know why my teacher uses the u,v coordinate system. I think I should ask him some day. Haha.

Ah yes, I got the triangle in the xy-plane as shown below:

Just wondering, how can I be very sure that I have placed the limits on my ∫ ∫ correctly? My integral equation is below:

Thank you! :)

Your limits are OK but I think you have u and v exchanged in the integrand. Another reason not to use u and v unnecessarily.

 

[galaxy_twirl]

I tried integrating the above and I got the same answer as my teacher, who did it in another way:

Hence, I think both methods work (or is this my method a 'fluke'? Haha.) Thanks. :)

 

[LCKurtz]

I tried integrating the above

No you didn't. In post #7 you wrote the integrand as $u-v+2$, which has the $u$ and $v$ switched as I said. Below you have the correct integrand.

I got the same answer as my teacher, who did it in another way:

Hence, I think both methods work (or is this my method a 'fluke'? Haha.) Thanks. :)

 

[galaxy_twirl]

No you didn't. In post #7 you wrote the integrand as $u-v+2$, which has the $u$ and $v$ switched as I said. Below you have the correct integrand.

I see. Haha. I think I misinterpreted your intended message. Sorry. Thanks for all your help! :D