Find the volume of the solid obtained by rotating the region in the first quadrant bounded by y=x^4 , y=1, and the y -axis about the line y=−5.

Answer 1

It's crucial here to draw a detailed picture. The region bounded by those three lines (y=x⁴, y=1, and the y-axis) is a wedge. It begins at x=0 and ends with a vertex at x=1.

To rotate this around y=5, imagine breaking the wedge down into vertical strips. The width of a strip is dx, the lower part of the strip is given by y=x⁴, and higher part is given by y=1.

If we rotate this strip about y=5, we will get a thin annulus whose inner radius is R₁=4 and R₂=5-x4. The width is, as before, dx. The volume of this thin annulus is given by dV=π*(R₂²-R₁²)*dx. If we plug in the expressions for R1 and R2, we get dV=π*((5-x⁴)²-16)*dx.

To get the full volume of the solid-of-revolution, we simply sum up all the volumes of the thin annuli, i.e. we integrate. Integrating dV from x=0 to x=1, we get V=(64*π)/9.