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Double Integral Working Help

Posted 9 years ago

Hello, I have no idea how show the working for this, it is supposed to calculate the surface area of a cone. I have the answer but i dont have an idea to approach the working.

Please Help!!!

Thank you

enter image description here

POSTED BY: Bob H
4 Replies

It would seem that your cone is 577.29 cm with a base of radius 28.25, Using the formula for the area of a cone,

A = Pi radius Sqrt[height^2+radius^2]

the area is 51295.8 cm^2. Something is still wrong.

Eric

POSTED BY: Eric Johnstone
Posted 9 years ago

Thanks, I was trying to find the surface area of a cone enter image description here through double integration and the radius of the cone is 28.25. So would the above be correct. Sorry I am only in high school so not really sure about some of this maths.

POSTED BY: Bob H

Are you sure the equation is written correctly? The two parenthesized terms are identical. This is what I get.

    In[10]:= NIntegrate[Sqrt[1+(577.29-(19.727)Sqrt[x^2+y^2])^2+
                               (577.29-(19.727)Sqrt[x^2+y^2])^2],{x,0,2 Pi},{y,0,28.25}]

    Out[10]= 71468.9

Eric

POSTED BY: Eric Johnstone
f[x_,y_]:=(*Type the given function here.*);
(* Numerically integrate with NIntegrate because on my Mma7, the function could not be integrated by Integrate*)
area = NIntegrate[ f[x,y], {x, 0, 2 Pi}, {y, 0, 28.25}]; 
POSTED BY: Isaac Abraham
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