The region bounded by `y=xsqrt(4-x^2)` and `y=0` revolved about the *x-axis* is shown on the attached image.We may apply the** Disk method** using a **rectangular strip perpendicular to the axis of revolution.** As shown on the attached image, the *thickness of the rectangular strip is "dx"* with a* vertical orientation perpendicular to the x-axis* (axis of revolution).

We follow the formula for the Disk method:`V = int_a^b A(x) dx ` where disk's base area is `A= pi r^2` with `r =y=f(x)` .

Note: *r = length of the rectangular strip*. We may apply `r = y_(above)-y_(below)` .

Then `r =(xsqrt(4-x^2))- 0 =xsqrt(4-x^2)` .

The boundary values of x are `a=-2` to `b=2` .

Plug-in the `f(x)` and the boundary values to integral formula, we get:

`V = int_(-2)^2 pi(xsqrt(4-x^2))^2 dx `

Simplify:

`V = int_(-2)^2 pix^2(4-x^2) dx`

`V = int_(-2)^2 pi*(4x^2-x^4) dx`

Apply basic integration property: `intc*f(x) dx = c int f(x) dx`

`V = pi int_(-2)^2 (4x^2-x^4) dx`

Apply basic integration property:`int (u-v)dx = int (u)dx-int (v)dx` .

`V = pi *[ int_(-2)^2 (4x^2)dx -int_(-2)^2(x^4) dx]`

Apply Power rule for integration: `int x^n dx= x^(n+1)/(n+1)` .

`V = pi *[(4x^3)/3 -x^5/5]|_(-2)^2`

Apply definite integration formula: `int_a^b f(y) dy= F(b)-F(a)` .

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

`V = pi *[32/3 - 32/5] -pi *[(-32)/3 -(-32)/5]`

`V = pi *[160/15 - 96/15] -pi *[(-160)/15 -(-96)/15]`

`V = pi *[64/15 ] -pi *[(-64)/15 ]`

`V =(64pi)/15 -( -64pi)/15`

`V =(64pi)/15 +64pi/15`

`V =(128pi)/15` or `26.81` (approximated value)

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