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The region between $y=x^2+1$ and $y=-x+3$ is rotated about the $x$-axis.

I have to compute the volume. The intersection between these two curves are at $x=-2$ and $x=1$.

At first, I thought about $V(x)=\pi\displaystyle\int_{-2}^1(3-x-(x^2+1))^2dx=\frac{81\pi}{10}$, where the radius is the difference between the two curves.

Another approach I made was $V(x)=\pi\displaystyle\int_{-2}^1[(3-x)^2-(x^2+1)^2]dx=\frac{117\pi}{5}$. I'm not sure why both results are different, and which one is correct.

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  • $\begingroup$ What's your reasoning for setting up the first integral? You have used the formula $V = \pi \int_a^b (R^2-r^2) \ dx$ incorrectly. $\endgroup$ – Toby Mak Jan 16 at 3:11
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    $\begingroup$ They disagree because $a^2-b^2\ne(a-b)^2$. $\endgroup$ – Brian M. Scott Jan 16 at 3:17
  • $\begingroup$ Yeah I just got it, thank you. $\endgroup$ – Fabrizio Gambelín Jan 16 at 3:19
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It might be easier to think about it in steps. So, we know the formula for the volume of a solid of revolution about the $x$ axis from $a$ to $b$, right?

$$V = \pi \int_a^b (f(x))^2 dx$$

Think of the $f^2$ as serving the role of a radius, and the integral as contributing the length by summing up over all of the super-small bits of length $dx$ on the interval.

If you have the area between two curves $f$ and $g$, though - let $f \ge g$ here - you can think of it not as revolving the area between $f$ and $g$, but rather as just revolving the outer area $f$ and then cutting a hole from it based on $g$, and you want the volume left over. That is, for instance,

$$V = \underbrace{\pi \int_a^b (f(x))^2 dx}_{\text{original volume}} - \underbrace{\pi \int_a^b (g(x))^2 dx}_{\text{hole you cut out}}$$

Simplifying, we get

$$V = \pi \int_a^b \Big( (f(x))^2 - (g(x))^2 \Big) dx$$

or, if you understand that $f^2(x)$ means $(f(x))^2$, you can make this a little neater to write:

$$V = \pi \int_a^b \Big( f^2(x) - g^2(x) \Big) dx$$

Thus your second approach would be the correct one.

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    $\begingroup$ I can clearly understand it now, thank you a lot. $\endgroup$ – Fabrizio Gambelín Jan 16 at 3:19

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