The region R enclosed by the coordinate axes and the graph of y=k(x-2)^2 is shown. When this region is revolved around the y-axis, the solid formed has a volume of 8 Pi cubic units. What is the value of k?
Set up the integral, according to the formulas you have learned, and obtain the variable-based answer. Set this answer equal to "8pi", and solve for k.
i seem to have some troubleright up to the actual integration. i think my limits may be off. right now after making the denomiantor the same for all terms i have 8 pi= 2 pi ((3kx^4-16kx^3+ 24kx^2)/12) from 4k to 0. Is that right?
You failed to include the diagram in your posting.
Guessing at what it might be, you can solve the following for k.
\(\displaystyle \L
\int_0^2 {2\pi x\left[ {k\left( {x - 2} \right)^2} \right]dx} = 8\pi\).
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