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\[\int\limits_{0}^{4}\int\limits_{0}^{y} \sqrt{9+y^2} dxdy\]
do i have to use trig substitution? if so, how?
So..you solve for x first, what do ya get?
i don't even know how to start, i find it hard to integrate the equation.
i'm thinking of substituting 3tan(theta) to y and then i got sec^3 theta, so i need to integrate this by parts. but it became so complicated.
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okay, you start integrating for x, so all your "y" expressions are constants and you have \[\int\limits_{0}^{4}\int\limits_{0}^{y}\sqrt{9+y^2}dxdy=\int\limits_{0}^{4}\sqrt{9+y^2}*x|(0toy)dy\] \[\int\limits_{0}^{4}\sqrt{9+y^2}(y-0)dy=\int\limits_{0}^{4}y\sqrt{9+y^2}dy\]
oh yes i forgot, y must be constant first! thanks!!!
i mean i must treat y as a constant*
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