Calculate the Volume of a Lemonsqueezer

[Elpinetos]

Homework Statement

f(x)=\frac{1}{81}*x^4-\frac{5}{9}*x^2+4
The tangent in Point P(6|0) when rotated around the y-Axis gives the Shape of the Squeezer. The bottom is at y=-5, the top at y=0

The Attempt at a Solution

First I calculated the tangent and got

t: y=4x-24
Then I converted that to x^2 since I need that for the rotation around the y-Axis
x^2=\frac{y^2}{16}+3y+36

Then I rotated it:

V=\pi\int\frac{y^2}{16}+3y+36dy = \pi(\frac{y^3}{48}+\frac{3y^2}{2}+36y)(0 to -5)
= \pi(0+0-(-\frac{125}{48}+\frac{75}{2}-180) = \pi(\frac{125}{48}-\frac{1800}{48}+\frac{8640}{48})
= \frac{6965\pi}{48} = 455.86

Since this is in cL, which would equal around 4.6L I find the result a bit excessive. Have I made a mistake somewhere? Can someone please check for me? Thank you in advance :)

 

[LCKurtz]

Homework Statement

f(x)=\frac{1}{81}*x^4-\frac{5}{9}*x^2+4
The tangent in Point P(6|0) when rotated around the y-Axis gives the Shape of the Squeezer. The bottom is at y=-5, the top at y=0

The Attempt at a Solution

First I calculated the tangent and got

t: y=4x-24
Then I converted that to x^2 since I need that for the rotation around the y-Axis
x^2=\frac{y^2}{16}+3y+36

Then I rotated it:

V=\pi\int\frac{y^2}{16}+3y+36dy = \pi(\frac{y^3}{48}+\frac{3y^2}{2}+36y)(0 to -5)
= \pi(0+0-(-\frac{125}{48}+\frac{75}{2}-180) = \pi(\frac{125}{48}-\frac{1800}{48}+\frac{8640}{48})
= \frac{6965\pi}{48} = 455.86

Since this is in cL, which would equal around 4.6L I find the result a bit excessive. Have I made a mistake somewhere? Can someone please check for me? Thank you in advance :)

No mistake. That is correct.

 

[Elpinetos]

4.6L Lemonsqueezer? Okay thank you^^

 

[LCKurtz]

4.6L Lemonsqueezer? Okay thank you^^

The problem states nothing about units. You can tell the 455.8 is reasonably close since the volume would be a bit less than a disk of radius $6$ and thickness $5$$$
5\pi 6^2 = 565.48$$

 

[Elpinetos]

Yeah it does. It states 1 unit = 1cm and the Volume in cl
Thank you though :)