volume of rotation: y = sqrt(x), y=2, x=0 about x = 4

[pamw]

Find the volume formed by the curves y = sqrt(x), y=2, and x=0, rotated around the line x = 4.

I got kind of a weird looking answer (224pi/15) so I'm thinking I did something wrong.

And I also have a more theoretical question. If I have take the same shape made by those three lines, but I rotate it around the y-axis or the x-axis, should I be getting the same volume?

Thanks!

 

[stapel]

I'm thinking I did something wrong.

If you reply showing your work (all your steps; not just the final answer), we'll be able to check and see where (and if) you made any errors.

If I have take the same shape made by those three lines, but I rotate it around the y-axis or the x-axis, should I be getting the same volume?

If you take a circle and rotate it about its own diameter, you'll get a sphere. If you take that same circle and rotate it about some coplanar line exterior to the circle, you'll get a "donut" shape. Obviously, since the sphere could fit inside the donut shape, the two volumes can't possibly be the same.

Now apply similar reasoning to your question.

Eliz.

 

[galactus]

I got kind of a weird looking answer (224pi/15) so I'm thinking I did something wrong.

That answer isn't weird. I believe it's correct.

Did you use shells?.

 

[skeeter]

washers w/r to y ...

\(\displaystyle \L V = \pi \int_0^2 16 - (4 - y^2)^2 dy\)

\(\displaystyle \L V = \pi \int_0^2 8y^2 - y^4 dy\)

\(\displaystyle \L V = \pi \left[\frac{8y^3}{3} - \frac{y^5}{5} \right]_0^2\)

\(\displaystyle \L V = \pi \left[\frac{64}{3} - \frac{32}{5} \right] = \frac{224 \pi}{15}\)