Question

Consider the region bounded to the right by the graph of x=ysin(3y)+3, to the left by the y axis, and above and below by y=65π and y=3π. Using the shell method, what is the volume of the solid of revolution obtained by rotating this region about the x-axis? Enter an exact value in terms of π.

Answer 1

The volume of the solid obtained by rotating the region between y=65π, y=3π, x=ysin(3y)+3 and the y-axis about the x-axis is found by integrating the product of the shell's height (ysin(3y)+3) and circumference (2πy) from y=3π to y=65π.

Step-by-step explanation:

The student is asking for the volume of the solid of revolution obtained by rotating the region bounded by the graph of x=ysin(3y)+3, the y-axis, and the horizontal lines y=65π and y=3π, about the x-axis using the shell method.

To apply the shell method, we consider a shell at a position y between 3π and 65π. The height of the shell is given by the x-value of the function, which is ysin(3y)+3, minus the x-value on the y-axis, which is 0, so the height will just be ysin(3y)+3. The circumference of the shell is 2πy, since it is the distance around the 'tube' created by rotating the shell around the x-axis. The volume of each shell is the product of its height, circumference, and thickness (dy). To find the total volume, we integrate this expression with respect to y from 3π to 65π:

∫3π65π [(ysin(3y) + 3) * (2πy) dy].

Through integration, we can then obtain the exact value of the volume in terms of π.