# Use the shell method to find the volume of the solid generated by revolving the plane region about the given line. `y = sqrt(x)` `y = 0, x = 4` about the line `x = 6` `y=sqrtx`

`y=0`

`x=4`

Axis of revolution `x=6`

Graph the given equations to determine the bounded region.

(Note: The blue curve is the graph of y=sqrtx. The green line is the graph of y=0. The red line is the graph of x=4. And, the yellow line is the graph of x=6.)

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`y=sqrtx`

`y=0`

`x=4`

Axis of revolution `x=6`

Graph the given equations to determine the bounded region.

(Note: The blue curve is the graph of y=sqrtx. The green line is the graph of y=0. The red line is the graph of x=4. And, the yellow line is the graph of x=6.)

Since the axis of revolution is vertical, when using shell method to determine the volume of thr solid formed, the formula is:

`V=int_a^b 2 pi r h dx`

where r  is the radius and h is the height of the cylimdrical shell.

Base in the graph above, the radius of the cylinder is the distance of the outer edge of the bounded region (which is the blue curve) from the yellow line.

`r=6-x`

And the height is the distance of the lower edge of the bounded region (green line) from the upper edge of the bounded region.

`h=y_U-y_L`

`h=sqrtx -0`

`h=sqrtx`

Moreover, the limits of the integral are the x-coordinates of the corners of the bounded region.

`a=0`

`b=4`

Plugging them  to the formula of volume, we would have:

`V=int_0^4 2pi (6-x)(sqrtx) dx`

Simplifying the integrand, it becomes:

`V=2pi int_0^4 (6sqrtx-xsqrtx)dx`

`V=2pi int_0^4 (6x^(1/2) -x^(3/2))dx`

Evaluating it yields:

`V=2pi (4x^(3/2) - 2/5x^(5/2))|_0^4`

`V=2pi((4*4^(3/2) -2/5*4^(5/2))-0)`

`V=2pi(4*(2^2)^(3/2)-2/5*(2^2)^(5/2))`

`V=2pi(4*2^3-2/5*2^5)`

`V=2pi(32-64/5)`

`V=2pi*96/5`

`V=(192pi)/5`

Therefore, the volume of the solid generated by revolving the bounded region at x=6 is  `(192pi)/5`    cubic units.

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