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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. sketch the region, the solid, and a typical disk or washer.

$y=2-\frac{1}{2}x$, $y=0$, $x=1$, $x=2$; rotate about the $x$-axis.

I tried using the formula $$ \int_1^2\pi\left(2-\frac{1}{2}x-0\right)^2\,dx $$ and got $27\pi/12$ but the answer was $32/3$... it seems like you would rotate the line about the x axis with this formula but it's not coming out right..

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  • $\begingroup$ "Answers" are not necessarily right. That said, I get neither your answer nor the "answer." $\endgroup$ Sep 3 '13 at 23:38
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$$\text{Region:}\;\; y=2-(1/2)x,\; y=0, \;x=1,\; x=2;\;\;\text{rotated about the x-axis}$$

Your procedure (approaching the problem) is correct. But I think both you and the "answer" are incorrect: perhaps a simple mistep in expanding the square, or a miscalculation. The answer you refer to may very well be a typo or an error.

$$\begin{align} \pi \int_1^2 \left(2 - \frac x2\right)^2\,dx & = \pi\int_1^2 \left(4 - 2x + \dfrac{x^2}{4} \right)\,dx \\ \\ & = \pi\Big[4x - x^2 + \dfrac{x^3}{12}\Big]_1^2 \\ \\ & = \pi\left[\left(8 - 4 + \frac{8}{12}\right) - \left( 4 - 1 + \frac{1}{12}\right)\right]\\ \\ & = \pi\left(1 + \dfrac 7{12} \right)\\ \\ & = \dfrac {19\pi}{12}\end{align}$$

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  • $\begingroup$ thats what I got as well but when I evaluated it I get 27pi/12.. and not 32/3... $\endgroup$
    – user92967
    Sep 3 '13 at 23:49
  • $\begingroup$ i see. i made a miscalculation and i was also looking at the wrong section for the answer. thanks for pointing these out.... $\endgroup$
    – user92967
    Sep 3 '13 at 23:51
  • $\begingroup$ You're welcome. Glad to help! $\endgroup$
    – amWhy
    Sep 3 '13 at 23:52
  • $\begingroup$ @amWhy: Also needs a TU! +1 $\endgroup$
    – Amzoti
    Sep 4 '13 at 0:53

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