The region between the graphs of x=y2 and x=5y is rotated around the line y=5. Find the volume of the resulting solid.

To find the volume of the resulting solid, we can use the method of cylindrical shells.

First, let's visualize the region between the graphs of x = y^2 and x = 5y. This is a parabolic region in the first quadrant bounded by the y-axis and the curve x = 5y.

We need to find the limits of integration for y to ensure that we cover the entire region. To do this, let's solve the two equations:

x = y^2 and x = 5y

By substituting x = y^2 into the second equation, we get:

y^2 = 5y

Rearranging this equation, we have:

y^2 - 5y = 0

Factoring out y, we get:

y(y - 5) = 0

So, y = 0 or y = 5.

Since we are rotating around the line y = 5, we need to shift the region up by 5 units. Thus, the new limits of integration for y become y = 0 + 5 = 5 and y = 5 + 5 = 10.

Now, let's consider a vertical slice of thickness Δy at a distance y from the line y = 5. This slice forms a cylindrical shell when rotated around the line y = 5.

The radius of the cylindrical shell at a given y-value is the distance between the line y = 5 and the curve x = y^2. This distance is given by x - 5 since the line y = 5 is our axis of rotation.

So, the radius of the cylindrical shell is r = x - 5 = y^2 - 5.

The height of the cylindrical shell is Δy since it is a vertical slice.

The volume of a cylindrical shell is given by the formula:

Shell volume = 2πrhΔy

where r is the radius, h is the height, and Δy is the thickness of the slice.

Therefore, the differential volume element of the resulting solid is:

dV = 2π(y^2 - 5)(Δy)

To find the volume of the solid, we need to sum up all these cylindrical shells. This can be done by integrating the differential volume element over the range of y.

V = ∫[5,10] (2π(y^2 - 5)) dy

Evaluating this integral gives us the volume of the resulting solid.