<s>[INST] Area of the surface generated by revolving curve around y-axis 
So I did something wrong in my solution because I'm not seeming to get the right answer.
$$\int_c^d 2\pi (4 \sqrt{9-y}\sqrt{1-\frac{4}{9-y}})~\mathrm{d}y$$ 
combine square roots and move out constants
$$\int_c^d 8\pi (\sqrt{5-y})~\mathrm{d}y$$
indefinite integral of $\sqrt{5-y}$ is $-\frac{2}{3}(5-y)^\frac{3}{2}$
move out constant $$-\frac{16}{3}\pi[(5-y)^\frac{3}{2} ]$$ from zero to $\frac{135}{16}$
$$-\frac{16}{3}\pi\left(-\frac{55}{16}^\frac{3}{2}-5^\frac{3}{2}\right)$$
 Anyway this is supposed to come out as $-\frac{73\pi \sqrt{73}}{12}+\frac{208\pi \sqrt{13}}{3}$ and I can't even fathom how to put it in that form.
 [/INST] $\sqrt{1-\dfrac{4}{9-y}}$ should be $\sqrt{1+\dfrac{4}{9-y}}$
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