Question help (1 Viewer)

unknown777 · Jul 9, 2009

Hey can anybody give me the solutions to this question.
The graph y=x^3-x^2-4x+10, bounded by the line y=3 and the region 0<=x<=3, is rotated about the x-axis. Find the volume formed.
Thanks

Drongoski · Jul 9, 2009

Assuming the curve is above the line y=3: the volume = vol of curve rotated about the x-axis minus the vol of the line y-3 rotated about the x-axis (for x from 0 t0 3)

$\100dpi \ \ =\ \ \int_{0}^{3}\pi y^{2}\ dx\ \ -\ \ \int_{0}^{3}\pi \times 3^{2}\ dx\\ \\ \ \ =\ \ \pi \int_{0}^{3}(x^{3}-x^{2}-4x+10)^{2}\ -\ 3^{2}\ \ dx\\ \\ =\ \ \pi \int_{0}^{3}x^{6}-2x^{5}-7x^{4}+28x^{3}-4x^{2}-80x+91\ dx\\ \\ =\ \ \pi \left [ \frac{1}{7}x^{7}-\frac{1}{3}x^{6}-\frac{7}{5}x^{5}+7x^{4}-\frac{4}{3}x^{3}-40x^{2}+91x \right ]^{3}_{0}\\ \\ =\ \ ?\ \ ?\ \ ?\ \ ?$

Edit

Didn't realise expression so messy; I could have made an algebraic fiddling error anywhere. So u or someone else work the final answer out.

unknown777 · Jul 9, 2009

Ans: (6063/35)pie

x.Exhaust.x · Jul 9, 2009

Drongoski · Jul 9, 2009

unknown777 said:
Ans: (6063/35)pie

Thankfully, if you subst x=3 (and x=0 giving only 0) you'll get the answer.

unknown777 · Jul 9, 2009

x.Exhaust.x said:
lol

Excuse me. What's so funny? Can't a human being ask a question?

x.Exhaust.x · Jul 9, 2009

unknown777 said:
Excuse me. What's so funny? Can't a human being ask a question?

You're human?

Question help (1 Viewer)

unknown777

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Drongoski

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unknown777

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x.Exhaust.x

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Drongoski

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unknown777

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x.Exhaust.x

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