Find the volume of the region of y=sqrt (x), y=2, and x=0 that revolves around y=2.

Question

The answer is 8pi/3. How do you find this?

Answer 1

Of course the proper approach is to make a sketch graph of the region in question. Its boundaries are the y-axis, the line y = 2, and part of the graph of sqrt(x), which intersects the line y = 2 at (4,2).  The region is triangularish, with the 'hypotenuse' being a concave-downward curve.

 

Now when this region revolves around the line y = 2, its cross-sections are circles of radius 2-√x.

 

We set up the definite integral (lower and upper x limits are 0 and 4) of integrand pi•(2-√x)²dx. **

 

Indeed, the answer is 8 pi / 3.  ** Note; the dx is the thickness of each circular disk cross-section.

Answer 2

Hi Pokenerd,

 

I wish I could draw this and be able to show you with a graph...

 

But here goes without a picture.  YOU should graph this so that you can follow what I am explaining.

 

When you graph it, you have a horizontal line at y=2, a vertical line at x=0, and the bottom right of the four regions created by this "+", has a curve that is concave down, starting at (0,0) and intersecting the line y=2 at the point (4,2).

 

Do you have this graphed?  It should look like a right triangle with a "curved in" hypotenuse.

 

 

Now, imagine putting a rod along the line y=2, and attaching the "triangle with a curved hypotenuse" along the horizontal line (y=2).  Now spin the rod really fast so that it looks like a solid is formed...  This is the solid we are trying to find the volume for.

 

If we could a disk out of this solid, parallel to the y-axis and perpendicular to the x-axis, and super skinny (dx in thiickness), it would have a volume of a cylinder ( V=(pi)r²h ) with a "r" of (2-sqrt(x)), and an "h" of "dx".  

 

Depending on what your x-coordinate is, your radius will change. The disk you would cut out when x=0 has a radius of 2 ([y=2] - [y=√0]), the disk you would cut out if x=1 have a radius of 1 ([y=2] - [y=√1]), the disk you would cut out if x=3 would have a radius of ([y=2]-[y=√3]).   All of these disks are infinitesimally "skinny" with a thickness (or cylinder height) of dx.  To add the volume of all of these really skinny disks, in Calculus, is INTEGRATION.

 

So, the integral you need is the sum of all of the skinny disks' volumes.

 

V = (pi)∫(2-sqrt(x))²dx  (from x=0 to x=4).

       (pi)      (r^2)     h

 

V = (pi) ∫ (4 - 4√x +x) dx  from x = 0 to x = 4

 

V = (pi) [ 4x - (8/3)x^(3/2) +(1/2)x² ]  evaluated at 4 and 1

 

V = (pi)( [ 4(4) - (8/3)(4)^(3/2) + (1/2)(4)² ]  - [4(0) - (8/3)0^(3/2) +(1/2)(0)²] )

 

V = (pi)(8/3)  or, as you wrote:   8pi/3