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Mathematics 10 Online
OpenStudy (anonymous):

integration master! I NEED HELP!

OpenStudy (anonymous):

\[\int\limits_{-3}^{3}\int\limits_{-\pi}^{\pi} \sqrt{1+\sin^2 (y) +x^2 \cos^2 (y)} dydx\]

OpenStudy (anonymous):

so this is finding surface area so original function was ( z=x sin y ) and limits is given

OpenStudy (anonymous):

so formula for the finding area is sqrt(1^2 + az/ax^2 + az/ay^2)

OpenStudy (anonymous):

hi can you see what i wrote?

OpenStudy (anonymous):

im reading it hold on....

OpenStudy (anonymous):

oo I love this

OpenStudy (anonymous):

hi

OpenStudy (anonymous):

haha i love this too

OpenStudy (anonymous):

I'm starting to think that maybe your set up was wrong

OpenStudy (anonymous):

ok so you know the formula for the finding area right?

OpenStudy (anonymous):

Its been a while

OpenStudy (anonymous):

ok so its

OpenStudy (anonymous):

\[\int\limits_{}^{}\int\limits_{D}^{} \sqrt{1^2 + (az/ax)^2 + (az/ay)^2} dA \]

OpenStudy (anonymous):

so z=x sin(y) so i got partial derivative sin(y) for X

OpenStudy (anonymous):

and i got partial derivative xcos(y) for Y

OpenStudy (anonymous):

and squared both X, Y -> plug in to the finding area formula

OpenStudy (anonymous):

have you tried switching the bounds?

OpenStudy (anonymous):

does bounds mean limits?

OpenStudy (anonymous):

its given by \[-3\le x \le3 , -\pi \le y \le \pi\]

OpenStudy (anonymous):

holy s*** your teachers are mean The thing I can think of is try switching all the trig functions to be the same then do some black magic substitutions to get an answer

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