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Question:

Find the volume of the solid formed by rotating completely about $x$-axis the area enclosed by a curve.

My answer:

I drew the curve and the area formed it is between $-2$ and $0$ (the rest of the curve after zero goes to infinite)

I calculated:

$$V=\pi\int_{-2}^0[x^3-2x^2]^2\,\mathrm dx=\pi\int_{-2}^0x^5-2x^4=\pi\int\dfrac{x^6}6-\dfrac{2x^5}5$$

I plug in the values (as shown bellow) and I get $-23\pi$ the correct answer should be $128\pi/105$. What am I doing wrong here? thank you

enter image description here

UPDATE: the region bound is actually between 2 and 0 not -2 and 0

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You did a very strange job! $$\int (x^3-2x^2)^2 \, \mathrm{d}x\neq\int(x^5-2x^4) \, \mathrm{d}x$$ Indeed: $$\int (x^3-2x^2)^2 \, \mathrm{d}x=\frac{1}{7}x^7-\frac{2}{3}x^6+\frac{4}{5}x^5+C$$ so $$V=\int_0^2\pi y^2 \, \mathrm{d}x=\pi \cdot \left(\frac{1}{7}2^7-\frac{2}{3}2^6+\frac{4}{5}2^5\right)-\pi(0)=\pi\times\frac{128}{105}$$

enter image description here

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  • $\begingroup$ should it be 4x^4 then $\endgroup$ – Cash Vai Dec 8 '13 at 13:42
  • $\begingroup$ Assuming the right integral for evaluating the volume is $\int_0^2 (x^3-2x^2)^2dx$ then you should focus to do $\int (x^6-4x^5+4x^4)dx$ instead. $\endgroup$ – Mikasa Dec 8 '13 at 13:45
  • $\begingroup$ how did you drive this? $\endgroup$ – Cash Vai Dec 8 '13 at 13:48
  • $\begingroup$ @CashVai: Form this (a-b)^2=a^2-2a*b+b^2. $\endgroup$ – Mikasa Dec 8 '13 at 14:10
  • $\begingroup$ oh thank you! second time making the same mistake! $\endgroup$ – Cash Vai Dec 8 '13 at 14:15

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