Shell Method: solid bounded by y = 3|x|, y = 3, abt. x-axis

[Shutterbug424]

Find volume of solid generated aeound the x axis and bound by the given curves:

y = 3 abs(x) ; y = 3

When I rationalize the problem using geometry, I get 9 pi.

It just doesn't seem right to me though.

Can anyone help?

 

[galactus]

Re: Shell Method

We can do y=3x and multiply by 2 because the of the symmetry involved with the absolute value..

\(\displaystyle 4{\pi}\int_{0}^{3}(\frac{y^{2}}{3})dy\)

Or with washers:

\(\displaystyle 18{\pi}\int_{0}^{1}(1-x^{2})dx\)

 

[Shutterbug424]

Thanks! This was really helpful. So I guess the answer in 12 pi.

Ok, here's a simple question. My brain is fried right now, so I can't remember:

How do you distribute the square in (3cost)^2 ?

 

[Deleted member 4993]

Thanks! This was really helpful. So I guess the answer in 12 pi.

Ok, here's a simple question. My brain is fried right now, so I can't remember:

How do you distribute the square in (3cost)^2 = 3[sup:11vexddj]2[/sup:11vexddj] * cos[sup:11vexddj]2[/sup:11vexddj]t?

 

[Shutterbug424]

Re: Shell Method

We can do y=3x and multiply by 2 because the of the symmetry involved with the absolute value..

\(\displaystyle 4{\pi}\int_{0}^{3}(\frac{y^{2}}{3})dy\)

Or with washers:

\(\displaystyle 18{\pi}\int_{0}^{1}(1-x^{2})dx\)

I found a formula for a right circular cone to be (pi * r^2 * h)/3. When I figure out the area of one side of the abs value graph I get that r=3 and h=1, therefore I get 3pi. Doubling that I get 6 pi. Why does this method give me a different answer? It seems like it should work. Any thoughts you have on this would be appreciated. Thanks