Question

Let R be the region in the first quadrant enclosed by the graph y=(square root(6x+4)), the line y=2x, and the y-axis.

a. Find the area of R.
b.Set up but do not integrate an integral,expression in terms of a single variable for the volume of the solid generated when R is revolved about the x-axis .
c. do the same as part b, just this time the y-axis.

Answer 1

Find the intersection points:
\[\sqrt{6x+4}=2x \implies 6x+4=4x^2 \implies \]
\[2x^2-3x-2= (2x+1)(x-2)=0, x=\frac{-1}{2}, x=2\]
Area:\[\int\limits_0^2 (\sqrt{6x+4}-2x)dx\]
Then around the x axis gives:
\[\pi \int\limits_0^2 ((\sqrt{6x+4})^2-(2x)^2)dx\]