Can someone confirm my answer?

[hank]

Ok, I have to find the volume of a solid that results when the region enclosed by the given curves is revolved around the x-axis.

y = 9 - x^2 and y = 0.

So I set it up this way:

V = pi S[0 - 3] [9 - x^2]^2 dx

= pi S[0 - 3] (81 - 2x^2 + x^4) dx
= pi [81x - (2/3)x^3 + (1/5)x^5] [ 0 - 3]
= pi(243 - 18 + (243/5)
= 1368pi / 5

Does this look right?

 

[pka]

Where you have \(\displaystyle {-2}x^2\) it should be \(\displaystyle {-18} x^2\)

 

[hank]

Ok, so after making that change I come up with 648pi / 5.

How's that?

 

[skeeter]

you found the volume if the region in quadrant I is rotated about the x-axis ... the original problem statement sez ...

... find the volume of a solid that results when the region enclosed by the given curves is revolved around the x-axis.

y = 9 - x^2 and y = 0.

the described region is in quadrant I and II ... double your solution.