Find the volume which is generated by revolving about the Y –axisFind the volume which is generated by revolving about the Y –axis the region in thefirst quadrant which is above the parabola y...

Find the volume which is generated by revolving about the Y –axis

Find the volume which is generated by revolving about the Y –axis the region in thefirst quadrant which is above the parabola y = x2and below the parabola y = 8 − x2.

 

Have it drawn but dont understand first quadrant.Need somebody to tell me how to move it on from here

Asked on by peterc1992

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to remember the formula of the volume of solid of revolution.

V = pi*`int_a^b(f(x))^2dx`

You need to determine the limits of integration a and b. To find a and b, you should solve the system of equations:

`y = x^2`

`y = 8 -x^2`

Equating the parabolas above, yields:

`x^2 = 8 - x^2`

Add `x^2`  both sides:

`2x^2 = 8 =gt x^2 = 4`  => `x_(1,2) = +-2`

You should keep only the positive value because you are restricted to the first quadrant.

The region that should be revolved about y axis: `y = 8 - 2x^2 =gt y - 8 = -2x^2 =gt 8 - y = 2x^2 =gt x = +sqrt ((8-y)/2)`

For `x = 2 =gt y = sqrt3`

The volume is:

V = `pi` `int_0^(sqrt3) (sqrt ((8-y)/2))^2dy`* => V = `pi*`  `int_0^(sqrt3)(8-y)/2dy` = `pi` (4`int_0^sqrt3 dy` - `int_0^sqrt 3(y/2)dy` )

`V = pi*(4sqrt 3 - 4*0 - 3/4 + 0/4)`

`V = pi*(16sqrt3 - 3)/4`

The volume of the solid of revolution is V = `pi*(16sqrt3 - 3)/4.`

bluemtnmath87's profile pic

bluemtnmath87 | High School Teacher | (Level 1) Adjunct Educator

Posted on

Don't let the "first quadrant" thing throw you off--that just means that you're only looking at the overlapping region inside the 1st quadrant.  Considering the graphs of these functions, it's clear that the lower bound of our integral will be at x = 0.  To find our upper bound, we simply need to find the parabolas' second point of intersection; that is, where x^2 = 8-x^2:

2x^2 = 8 --> x^2 = 4 --> x^2 = +/- 2.  Again, in the first quadrant, so only x=2 matters.  So our lower bound is x = 0 and our upper bound is x= 2.

So remember that to find the area, just find

2pi*INT[0,2](8-x^2 - x^2)xdx  (taking the integral of the area by upper function - lower function)

= 2pi*INT[0,2](8x-2x^3)dx  (after some simplifying)

I'll let you handle the rest! =D

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