Volume - Calculus

Question

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.

Answer 1

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

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

Answer 2

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.