find the volume of the solid generated by revolving the region about the given line. the region in the first quadrant bound above by the line y=1, below by the curve y=√(sin5x) ,and on the left by the y-axis, above the line y=-1
3 answers
What is the "given line" that the solid volume revolves around?
Should the word "above" in the last line be "about"? Then the question would make sense.
using washers, the volume is
∫[0,π/10] π R^2-r^2 dx
where R=2 and r = √(sin5x) + 1
v = ∫[0,π/10] π (4 - (√(sin5x)+1)^2) dx
I think you are in trouble.
∫√(sin5x) dx
cannot be evaluated using elementary functions.
∫[0,π/10] π R^2-r^2 dx
where R=2 and r = √(sin5x) + 1
v = ∫[0,π/10] π (4 - (√(sin5x)+1)^2) dx
I think you are in trouble.
∫√(sin5x) dx
cannot be evaluated using elementary functions.
0 answers