Use the shell method to find the volume generated by revolving the region bounded by y=6x-5, y=square root of x, and x=0 about the y-axis.  the volume is____ cubic units.
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The curve y=sqrt(x) and line y=6x-5 intersect where sqrt(x)=6x-5. Solve this by squaring: x=36x^2-60x+25 which becomes 36x^2-61x+25=0 which factorises (36x-25)(x-1)=0, from which x=1 or 25/36. At these values y=1 or -5/6. However the point (25/36,-5/6) is below the x axis and therefore below the bounds addressed in the question where (0,0) is the lower bound.
The purpose of the line appears to be to establish the upper bound of the curve (1,1); it isn't involved in the actual volume of rotation. Consider a hollow cylinder of thickness dx with the y axis as its central axis. The radius of the cylinder is x and its height is y=sqrt(x). The circumference is 2(pi)x so the volume of the cylinder is 2(pi)xsqrt(x)dx=2(pi)x^3/2dx. The integral of this is 2(pi)(2x^5/2)/5=(4(pi)x^5/2)/5. When we apply the limits 0 and 1 we get 4/5(pi)=2.513 cubic units approx.

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