Question

Why do we set z=0 to get the limits for the double integral in 2x+2y+z=8?

Answer 1

\[\int\limits_{}^{}\int\limits_{R}^{}\sqrt{\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}+1}dA\]
\(\Large \int_{0}^{4}\int_{0}^{4-x}3dA\)

Answer 2

Question #2 in the attachment.

Answer 3

It says in the first octant.

Answer 4

@IrishBoy123

Answer 5

\(\Large \int_{0}^{4}\int_{0}^{4-x}3 ~dy~dx\)

Answer 6

zela

i thought the 2 were connected. give me a second.

Answer 7

I do know that, but why do you think the limit for dy is y=4-x?

Answer 8

for the intersection of the plane 2x +2y +z = 8 and the xy plane, you set z to 0, so you have the line 2x + 2y + 0 = 8, ie x + y = 4. or y = 4 - x, or x = 4 - y.

that is the limit of the projection of the plane in the 1st octant.

if you were sitting in the crow's nest, so to speak, ie looking directly down from above the xy plane, you could describe that projection in several ways.

you could describe it as the combination of

0 < y < 4 - x and 0 < x < 4

or

0 < x < 4 - y and 0 < x < 4

the thing is that you are doing an iterated integral.

Answer 9

PS having added in the squares to your posted formula, ie

\[\int\limits_{}^{}\int\limits_{R}^{}\sqrt{\left(\frac{\partial z}{\partial x}\right)\color{red}{^2}+\left(\frac{\partial z}{\partial y}\right)\color{red}{^2}+1} \quad dA\]

i realise that what i have found somewhat confusing is that, whilst this is similar to the formula given on Paul's online, i would see it more intuitive to have either:

\[\int\limits_{}^{}\int\limits_{R}^{} \; \quad dA\] where \(dA = \sqrt{\left(\frac{\partial z}{\partial x}\right)\color{red}{^2}+\left(\frac{\partial z}{\partial y}\right)\color{red}{^2}+1} \quad dx \; dy\)

or:

\[\int\limits_{}^{}\int\limits_{R}^{}\sqrt{\left(\frac{\partial z}{\partial x}\right)\color{red}{^2}+\left(\frac{\partial z}{\partial y}\right)\color{red}{^2}+1} \quad dx \; dy\]

this might seem like nit picking or superficial....but the dA is an area element that should always IMHO take its meaning from the circumstances/context, it is not just shorthand for dx dy.

so if you were looking to calculate the volume under the plane in the 1st octant, you would be looking at \[\int\limits_{}^{}\int\limits_{R}^{} \; z \quad dA\]


if you don't mind, i'll tag @Phi who knows this stuff inside out.

note also that there are other, IMHO superior ways to do these. from the attachment, in my opinion, all you ever need to know to do these integrals is this very easy to remember idea:

Answer 10

Thank you!!!!!!!!!!!! It makes a lot of sense. @IrishBoy123 You are awesome :D

Answer 11

that's v kind zela, really glad it helped you

still hoping phi swings by, even after the fact.

good luck!

Answer 12

Thanks! :)