Calculating Volume of Region Revolved About y=4

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Find the region enclosed by x=3y and x=-y^2+4. Set up integrals both shell and disc that represent the volume when this region is revolved about y=4.

i got the shell method, how would i represent the disc method>

pi integral(-4,1) (3y-4)^2-(-y^2)^2 --this is wat i got for disc method, though i am sure it is wrong...
 
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It might help you if you make a drawing -- draw the line that you're rotating about, and the region that is being rotated.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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