Definite Integral Question (1 Viewer)

jash1228 · Dec 21, 2016

Find the volume of the solid of revolution formed when the curve y= x(x^3-3)^2 is rotated about the x axis from x=0 to x=1.

Answer: 211/15 (pi)^3 units

I keep getting close to the answer but don't know where I made the mistake in the workings. Thanks in advance!

InteGrand · Dec 21, 2016

jash1228 said:
Find the volume of the solid of revolution formed when the curve y= x(x^3-3)^2 is rotated about the x axis from x=0 to x=1.

Answer: 211/15 (pi)^3 units

I keep getting close to the answer but don't know where I made the mistake in the workings. Thanks in advance!

$\noindent The answer is $V = \pi\int_{0}^{1}x^{2}\left(x^{3} -3\right)^{4}\,\mathrm{d}x$. We can substitute $u=x^{3}-3$, $\frac{1}{3}\mathrm{d}u = x^{2}\,\mathrm{d}x$. When $x=0$, $u=-3$, and when $x = 1$, $u=-2$. So$

$\begin{align*}V &= \pi \int_{-3}^{-2}\frac{1}{3}u^{4}\,\mathrm{d}u \\ &= \frac{\pi}{3}\times \left[\frac{1}{5}u^{5}\right]^{-2}_{-3} \\ &= \frac{\pi}{15}\left(-2^{5} + 3^{5}\right) \\ &= \frac{211\pi}{15}.\end{align*}$

In other words, there's no pi-cubed in the answer, it's just pi. Maybe the answers meant to say cubic units, or just did a typo.

Definite Integral Question (1 Viewer)

jash1228

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InteGrand

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