Another volume of a curve question

Question

legomyego180 · Answer

$$x=\sqrt{4-y^2}$$

x=0

legomyego180 · Answer

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legomyego180 · Answer

So from this point im confused. Do I only flip the area in quadrant I?

phi · Answer

what are the bounds on this volume? is it the x=0 line ?

if so, you can do washers, with radius r= y  (but in terms of x) and integrate over x

legomyego180 · Answer

yup x=0

so

$$\int\limits_{-2}^{2}\pi(\sqrt{4-y^2})^2-\int\limits_{-2}^{2}\pi(0)^2$$
?

phi · Answer

and what are we rotating around.  I was guessing the x-axis  (i.e. y=0) ?

legomyego180 · Answer

rotating around the y-axis

legomyego180 · Answer

Sorry bounds are -2, 2

phi · Answer

the shape starts at x=0 and goes to x=2 
(you said x=0 is the lower bound, right ?)

legomyego180 · Answer

In Exercises 27-44 , sketch the region Ω bounded by the curves. Find the volume of the
solid generated by revolving Ω about the y − axis.

38.) $$x=\sqrt{4-y^2}$$$$x=0$$

phi · Answer

ok, I sketched the figure.

legomyego180 · Answer

Bounds would be with respect to y from -2 to 2 I think, because we are revolving it around the y axis

legomyego180 · Answer

I think it would be:
$$\int\limits_{-2}^{2}\sqrt{4-y^2}$$

phi · Answer

ok, around the y-axis.  So my picture was wrong.  But what is x=0 ?  don't they mean y=0 for the bounds?

phi · Answer

ok I'll go with your theory
but you want pi r^2   where r is the x value (in terms of y)

legomyego180 · Answer

Im not sure, x=0 is just the y-axis so its kind of implied in the question that we would be rotating arounf that anyway? Im not sure why its there.

legomyego180 · Answer

Oh yea, forgot. Okay.

legomyego180 · Answer

Alright sweet @phi I think I got it. Thanks for your help