# Use the shell method to find the volume of the solid generated by revolving the region bounded by the given curves and lines about the y-axis.y=5x, y=-x/5, x=1   The answer is 52pi/15, can anyone...

Use the shell method to find the volume of the solid generated by revolving the region bounded by the given curves and lines about the y-axis.

y=5x, y=-x/5, x=1

The answer is 52pi/15, can anyone show me the process?

sciencesolve | Certified Educator

You need to find the point of intersection of curves y = 5x and y = -x/5.

Solving the system of equations y = 5x and y = -x/5 yields: 5x = -x/5 => 5x + x/5 = 0 => 24x + x = 0 => 26x = 0 => x = 0.

The next point of intersection is given: x = 1.

Checking what function is greater than the other on the interval [0,1] yields 5x > -x/5 => 5x + x/5 > 0

You need to use the formula of surface area of cylindrical shell to find the volume of the solid: V = `2pi*r*h`  (r is the distance from y axis of rotation => r=x and h denotes the height of cylinder = 5x + x/5.

You need to integrate the function `2pi*x*(5x + x/5).`  `int_0^1 2pi*x*(5x + x/5)dx = 2pi*int_(0^1)(5x^2 + x^2/5)dx` `int_0^1 2pi*x*(5x + x/5)dx = 2pi*(5x^3/3 + x^3/15)(0-gt1)` `int_0^1 2pi*x*(5x + x/5)dx = (2pi*(25x^3+x^3))/15 (0-gt1)` `int_0^1 2pi*x*(5x + x/5)dx = (2pi*26x^3)/15(0-gt1)` `int_0^1 2pi*x*(5x + x/5)dx = (52pi/15)*(1^3 - 0^3)`

`int_0^1 2pi*x*(5x + x/5)dx = (52pi/15)`

The volume of the solid of revolution is of (`52pi/15` ).