ques

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

For any level surface S given by \[\phi(x,y,z)=c\] Then the region R that is it's projection on xy plane is always given by \[\phi(x,y,0)=c\] ?
That seems true .
Is this a true false question?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Nope, it's a doubt that I have
we can do an example
Idk, can you think of a counter example??
can't think of a counterexample
hmmm let's ask someone else @IrishBoy123
maybe he can tell :P
let's test it first with a simple surface, a plane, which i shall just make uo \[\pi: 2x + 3y + 4z = 5\] \[\phi(x,y,z) = 2x + 3y + 4z [= const]\] \[\phi(x,y,0) = 2x + 3y =5\] you might think of this in terms of the line of intersection of plane \(\pi\) and the xy plane. in practise the projection is everywhere. the projection can be found from the surface integral formulation \[\iint_{R} \frac{1}{|\hat n \bullet \hat k|} \ dx \ dy\] \[= \frac{1}{4} \iint_{R} \ dx \ dy\]
maybe we need to understand what projection means "intersection"? then yes. just set z to zero for intersection projection, no
another example paraboloid \(z = x^2 + y^2 - 4\) \(\phi(x,y,z) = x^2 + y^2 - z = const \ \ [ie \ 4]\) \(\phi(x,y,0) = x^2 + y^2 = 4\) that's the circle on the xy plane, ie the *intersection*
\[\frac{1}{4}\int\limits_{0}^{\frac{2}{5}} \int\limits_{0}^{\frac{5-2x}{3}}dydx=\frac{1}{12}\int\limits_{0}^{\frac{2}{5}}(5-2x)dx=\frac{1}{12}[5x-x^2]_{0}^{\frac{2}{5}}\] Yeh sorry my bad, what I mean is that the equation we can form from \[\phi(x,y,0)\] is what we will use as a limit when we transform from S to R
What I'm trying to say is something like \[\phi(x,y,0)\] will always give a curve in xy plane that bounds R
that looks good to me!
i think that's right
So there's no reason to draw a graph every single time, we can just set z=0 and get our curve's equation, then we can double integrate within the limits
well, i always draw. always. seeing is believing. thing is you can get the line of intersection but you might still want to know if it is above or below the xy plane loads of people on here don't draw and they seem fine. to me, it's madness :-) and by draw, i mean rough sketch with key numbers. i use drawings to get the limits right, for example. you know, switch you integral from dydx to dxdy? i'll do that with my sketch. the integral http://www.wolframalpha.com/input/?i=%5Cfrac%7B1%7D%7B4%7D%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B2%7D%7B5%7D%7D+%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B5-2x%7D%7B3%7D%7Ddydx
Good insight, this time I'll have less questions for my exams because it's university exams, in schools we had like 30 questions, but this time there will be around 7 questions of high weightage and each question would require more depth and explanation, so it might just help to draw a graph actually !!
ah! 2/5 in limit should be 5/2 ?!?! i spotted that from my sketch :-))
Oops, \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] Not \[ax\]!!
25/48
Once again thanks!!
good thread!!

Not the answer you are looking for?

Search for more explanations.

Ask your own question