Volume around the x/y axis query? (1 Viewer)

[kloudsurfer]

Hey,
I was having some trouble with this question:

Find the volume of the solid of a revolution formed when the curve x^2 + y^2=9 is rotated about the x-axis between x=1 and x=3

I kept getting it wrong because i was going

x^2 +y^2 = 9
y = √(9-x^2)

y=3-x

And then I would go on to find the volume (cant be bothered to write it out here lol)...

Anyway, eventually i figured out that I shouldnt have done that last step (in bold), and thats why i kept getting a wrong answer. I now know how to do it and get the right answer, but I still dont know why what i was doing originally doesnt work.

Can someone please explain to me why you cant do that?

Thanks

 

[-pari-]

can you put up what the answer is meant to be?

just so i know whether i'm even right or not b4 i post it up

 

[kloudsurfer]

can you put up what the answer is meant to be?

just so i know whether i'm even right or not b4 i post it up

Oh right lol.

The answer is 28/3pi or 29.32 units^3

 

[-pari-]

I now know how to do it and get the right answer, but I still dont know why what i was doing originally doesnt work.

Can someone please explain to me why you cant do that?

o right you dont need the answer...

ok well

rt(9 - x^2) =/= 3 - x

because:

think of it this way. replace x by ...for eg, 2
here... (9 - x^2) = (9 - 4) =/= (3 - x)

you always solve the brackets first. but here, you dont know the value of x, so you can't solve whats in the brackets - rt(9 - x^2) is the simplest form.

did that make sense? :S

 

[ssglain]

√(9-x^2) =/= 3-x
because 9 - x^2 =/= (3-x)^2

The square root of a difference and the difference between square roots are not interchangeable. You cannot rewrite √(a - b) as √a - √b.

For that particular question, remember that
V = pi * int y^2 dx [lim: a-->b]

There is no need to square root both sides for an expression for y. All you need to do is make y^2 the subject in the given equation to obtain:
y^2 = 9 - x^2

So, V = pi * int (9 - x^2) dx [lim: 1-->3]
V = pi * [9x - (x^3)/3] with [lim: 1-->3]

I will leave the rest for you to solve.

 

[kloudsurfer]

√(9-x^2) =/= 3-x
because 9 - x^2 =/= (3-x)^2

The square root of a difference and the difference between square roots are not interchangeable. You cannot rewrite √(a - b) as √a - √b.

Ohhhhh. That makes sense. I always make that same mistake with different kinds of questions.

Thanks everyone.

 

[Horseypie]

You could always consider the graph of the curve, and knowing it will turn out to be a sphere of radius 3, use the equation V=4/3(pi*r^3)...without directly integrating.