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## Homework Statement

[itex]f(x)=\frac{1}{81}*x^4-\frac{5}{9}*x^2+4[/itex]

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

[itex]t: y=4x-24[/itex]

Then I converted that to [itex]x^2[/itex] since I need that for the rotation around the y-Axis

[itex]x^2=\frac{y^2}{16}+3y+36[/itex]

Then I rotated it:

[itex]V=\pi\int\frac{y^2}{16}+3y+36dy = \pi(\frac{y^3}{48}+\frac{3y^2}{2}+36y)(0 to -5)[/itex]

[itex] = \pi(0+0-(-\frac{125}{48}+\frac{75}{2}-180) = \pi(\frac{125}{48}-\frac{1800}{48}+\frac{8640}{48})[/itex]

[itex] = \frac{6965\pi}{48} = 455.86[/itex]

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 :)

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