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I need some help on this question: Find the volume of the solid formed when the region bounded by the curves $y = x^3 + 1, x = 1$ and $y = 0$ is rotated about the $x$-axis.

I did $$V = \pi \int_{-1}^0 (x^3+1)^2 dx = \frac{23\pi}{14},$$ but that was wrong. Any help would be great.

Thanks

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    $\begingroup$ Check your limits of integration. One of them is incorrect. Other than that, you should be OK. Also, check your algebra and/or your arithmetic. For the integral you did, even with the limits you used, it doesn't come out to $\frac{23\pi}{14}$. $\endgroup$ – quasi Mar 13 '17 at 16:43
  • $\begingroup$ Camryn, welcome to Math.SE! I hope you stick around to contribute to the site. $\endgroup$ – gt6989b Mar 13 '17 at 16:44
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Hint: Try integrating $$V = \pi \int_{-1}^1 (x^3+1)^2 dx$$

(You're given $x=1$ which will be your upper limit of the integral.)

You can see the region enclosed by the curve and the two given lines x=1 (vertical) and $y=0$ (the x-axis.)

enter image description here

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  • $\begingroup$ You're Welcome, camryn. $\endgroup$ – amWhy Mar 13 '17 at 17:53
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Hint: You have the right idea, except that $y = x^3 + 1$ is not an even function. Hence, the integral over $[-1,0]$ isn't the same as over $[0,1]$.

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