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I am having trouble with this question:

Find a formula for the volume of the solid obtained by rotating the area between the curve $y = \smash{\frac{1}{\sqrt{x}}}$ and the $x$-axis between $x=a$ and $x=b$ $(0<a<b)$ about the line $y = -1$.

I originally though that we would have the radius being $\smash{\frac{1}{\sqrt{x}}} + 1 $ and we would use $$ V = \pi \int_a^b r^2\, \mathrm{d}x $$ type of formula but that hasn't worked it seems.

Could someone show me where this is going wrong? Thank you.

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2 Answers 2

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You have the right idea, but you should sketch a picture of a typical cross-sectional region. For each $x$ in the interval $[a, b]$, the region will be an annulus (a circular disk minus a concentric smaller disk) whose area is $$ A(x) = \pi (R^2 - r^2), $$ where $R = 1 + x^{-1/2}$ and $r = 1$.

The area $A(x)$ is given by the formula (try to work this out yourself before revealing):

$$ A(x) = \pi \bigl( (1 + x^{-1/2})^2 - 1^2 \bigr) = \pi \bigl( x^{-1} + 2x^{-1/2} \bigr). $$

Thus, the volume is

$$ \begin{align} V &= \int_a^b A(x)\, \mathrm{d}x \\ &= \pi \int_a^b \bigl( x^{-1} + 2x^{-1/2} \bigr) \, \mathrm{d}x \\ &= \Bigl. \pi \bigl( \ln x + 4x^{1/2} \bigr) \Bigr|_a^b \\ &= \pi \bigl( \ln \tfrac{b}{a} + 4(\sqrt{b} - \sqrt{a}\,) \bigr). \end{align} $$

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  • $\begingroup$ Yes, I had an idea similar to this. This makes a lot of sense. Do you have any advice when it comes to trying to picture these types of questions? I think getting that down would tremendously help. Thank you for your assistance. $\endgroup$
    – MSM
    Feb 8 at 17:19
  • $\begingroup$ Any time you're using integration to calculate a geometric quantity (volume, area, surface area, etc.) think of how the pages of a book make up the book. You have to identify the axis along which you're integrating, and for each particular point in the interval you have to identify what in particular is being accumulated. $\endgroup$ Feb 21 at 17:21
  • $\begingroup$ When it's a solid of revolution and you're integrating over an interval parallel to the axis of revolution, pick an arbitrary point and work out what the cross section will look like. Often this can be drawn "one dimension down", i.e. in this problem, you can sketch a picture in the $(x, r)$-plane, where for each $x$ satisfying $a \leq x \leq b$, there's an interval in the radial direction from the inner to outer curve: $1 \leq r \leq 1 + \smash{\sqrt{x}}$. Spinning this interval around the the axis $r = 0$ clearly sweeps out an annulus. $\endgroup$ Feb 21 at 17:24
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I'm focusing on the phrasing on the question: ".... the area between the curve $y = \frac{1}{\sqrt(x)}$ and the x-axis....". Not the line $ y = -1$. This shape would have a hollow inner tube between $ y = -1 $ and the x-axis. This shape is formed by stacking infinite rings, not circles. Try subtracting this volume from your integral.

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