Is the Integral of sqrt(x^2 + y^2 + C^2) Solvable?

[Kruger]

I attached this integran (see picture). If I put this in my calculater he doesn't give me a formula independant of other integrals. Is this unsolveable? For you informations, C is a constant.

 

[steven187]

hello there

in your diagram the intergrand is not being integrated dy and dx should be placed after
the intergrand, correct me if I am wrong

steven

 

[saltydog]

That's a common notation especially when the integrals are nested (folded) deep. It's just:

\int\int \sqrt{x^2+y^2+C}dydx

 

[Zurtex]

Do you know how to integrate:

\int \sqrt{t^2 + c} \, dt

If you can do that the rest is pretty simple.

 

[Kruger]

no, sorry I don't know how to integrate this. Can someone tell me please?

 

[steven187]

hello there

just make a substitution which forms a change of variable such that you get something much more simpler in which you are able to integrate, as for your double integral, once you have put the intergrand in the correct position you first integrate the inside then you will be left with one integral and then integrate the next, but for each integral you are only integrating with respect to one variable and so all other variables are treated as constants in the intergrand

take care

steven

 

[Kruger]

Ok, I'll try it.

 

[geniusprahar_21]

integrate it by the lebniz product rule. take one function as sqrt(t2 + c) and the other as one.

 

[dextercioby]

Yes,it's reccomendable to use part integration,because the sign of the unknown constant can be either plus or minus and one wouldn't know which sub (cosh or sinh) to make.

Daniel.

 

[rachmaninoff]

That's a common notation especially when the integrals are nested (folded) deep. It's just:

\int\int \sqrt{x^2+y^2+C}dy \; dx

You left of a square, it was actually

\int\int \sqrt{x^2+y^2+C^2}dydx

which makes it easier (C^2 is nonzero).