Finding volume using integrals

[donjt81]

So here is the question

find the volume of the solid generated by revolving the region bounded by the curve y = sqrt(16 - x^2) and the line y = 0 about the x axis.

this is how I solved it

$$\int_{0}^{4} \Pi (16 - y^{2}) \; dy$$
$$\Pi \int_{0}^{4} (16 - y^{2}) \; dy$$

pi(16*y - y^3/3) from 0 to 4
pi(16*4 - 64/3 - 0)

so the answer I got is (128 pi)/3

is this correct.

 

[radou]

Unless I'm missing something, you should multiply your whole integral with 2, since your function is defined from -4 to 4 and symmetric.

 

[arildno]

Well, why haven't you used -4 as your lower limit?
Note that this would double the volume of your object.

 

[HallsofIvy]

Another check is to recognize that $y= \sqrt{16- x^2}$ is the upper half of a circle and so the figure formed is a sphere of radius 4. Its volume is $(4/3)\pi (4)^3$, twice your answer.

 

[donjt81]

thanks guys i completely missed that. you are right it should be -4 to 4... other than that does everything else look ok.