`y = 4 - x^2` Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.

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marizi eNotes educator| Certified Educator

To find the volume of a solid by revolving the graph of y =4-x^2 about the x-axis, we consider  the bounded region in between the graph and the x-axis. To evaluate this, we apply Disk method  by using a rectangular strip  perpendicular to the axis of rotation. As shown on the attached image, we consider a vertical rectangular strip with a thickness =dx

We follow the formula for  the Disk Method in a form of: `V = int_a^b pir^2 dx` or `V = pi int_a^b r^2 dx`

 where r is the length of the rectangular strip.

 In this problem, we let the length of the rectangular strip`=y_(above)-y_(below)` .

 Then` r = (4-x^2) - 0 = 4-x^2`

Boundary values of x: `a= -2` to `b=2` .

Plug-in the values on the formula `V = pi int_a^b r^2 dx` , we get:

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

Expand using FOIL method:`(4-x^2)^2 = (4-x^2)(4-x^2)= 16-8x^2+x^4` .

 The integral becomes:

`V =pi int_(-2)^2 (16-8x^2+x^4) dx`

Apply basic integration property:`int (u+-v+-w)dx = int (u)dx+-int (v)dx+-int(w)dx`  to be able to integrate them separately using Power rule for integration:  `int x^n dx = x^(n+1)/(n+1)` .

`V = pi[ int_(-2)^2(16) dx -int_(-2)^2(8x^2)dx+int_(-2)^2(x^4)dx]`

`V = pi[16x-(8x^3)/3+x^5/5]|_(-2)^2`

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

`V = pi[16(2)-(8(2)^3)/3+(2)^5/5]-pi[16(-2)-(8(-2)^3)/3+(-2)^5/5]`

`V =pi[32-64/3+32/5]-pi[-32-(-64)/3+(-32)/5]`

`V =pi[32-64/3+32/5]-pi[-32+64/3-32/5]`

`V=(256pi)/15 -(-256pi)/15`

`V=(256pi)/15 +(256pi)/15`

`V=(512pi)/15 ` or `107.23` (approximated value)

 

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