Finding the surface area of a balloon

This is the problem I'm working on. I think I have everything written out correctly, but I'm having trouble putting it all together.

Answer 1

[Voss]

surfaceBalloon(1,[1:3]) % check consistency with the prompt

ans = 1×3

5.0004 5.1214 5.3785

M = linspace(0,6,100); % calculate and generate plot
plot(M,surfaceBalloon(14000,M));

function A = surfaceBalloon(V,M)
% V = pi*R^3*(2+M)/3
R = (3*V/pi./(2+M)).^(1/3); % calculate R first
A = pi*R.^2.*(2+sqrt(1+M.^2)); % then calculate A
end

[Olivia Rose]

If I use the calculation for R that you have it comes back with an even smaller, and more incorrect, answer than before

[Voss]

I don't know about that. My answer seems to match the prompt and the plot.

Can you post the function you're using?

[Torsten]

But as shown, _'s code reproduces the M-A-curve given in your assignment. And the formulae are correct.

What do you mean by "it comes back with an even smaller, and more incorrect, answer than before" ?

[Olivia Rose]

The function I used is included in the original question.

And it has to have an output argument assigned.

[Torsten]

Yes, the correct function you should use is

function surfaceArea = surfaceBalloon(V,M)
% V = pi*R^3*(2+M)/3
R = (3*V/pi./(2+M)).^(1/3); % calculate R first
surfaceArea = pi*R.^2.*(2+sqrt(1+M.^2)); % then calculate A
end

instead of yours.

The task was to return the surface area for an arbitrary volume. You assume V = 14000, I guess.

I did not check whether the other calculations are ok.

[Voss]

@Danielle Reis I see. I thought maybe you had modified your function based on the answers you've seen here.

The function I posted has an output argument, called A in my version (called surfaceArea in @Torsten's version).

In my answer, I use the output from the function directly in plot(); it is not assigned to a variable in the calling workspace. Of course you can assign its output to a variable if that is a requirement.