The region between the graphs of x=y^2 and x=4y is rotated around the line y=4.
The volume of the resulting solid is
how so i solve?
To find the volume of the solid formed by rotating the region between the graphs of x=y^2 and x=4y around the line y=4, you can use the method of cylindrical shells. Here are the steps to solve it:
1. First, find the points of intersection between the two curves. Set the equations x=y^2 and x=4y equal to each other and solve for y:
y^2 = 4y
y^2 - 4y = 0
y(y - 4) = 0
So, y=0 and y=4 are the points of intersection.
2. Next, determine the limits of integration. Since we are rotating around the line y=4, the vertical strips will have a height from y=0 to y=4. So, the integral for the volume will be from y=0 to y=4.
3. Now, set up the integral for the volume using the formula for the volume of a cylindrical shell:
V = ∫[from y=0 to y=4] 2πrh dy
The radius, r, is the distance from the y-axis to the curve x=y^2. So, r = y^2. The height, h, is the difference between y and 4. So, h = y - 4.
4. Substitute the values of r and h into the integral:
V = ∫[from y=0 to y=4] 2π(y^2)(y - 4) dy
5. Integrate the expression with respect to y:
V = 2π ∫[from y=0 to y=4] (y^3 - 4y^2) dy
V = 2π [ (1/4)y^4 - (4/3)y^3 ] evaluated from y=0 to y=4
V = 2π [ ((1/4)(4^4) - (4/3)(4^3)) - ((1/4)(0^4) - (4/3)(0^3)) ]
V = 2π [ (64/4 - 64) - (0 - 0) ] = 2π (16 - 16) = 0
Therefore, the volume of the resulting solid is 0.