1
$\begingroup$

I'm trying to calculate the volume between the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and the line $x = 2a$ around the $y$ axis using two methods but I'm getting different answers:

  1. Using volume of a solid of revolution with circular-ring method (the usual formula is multiplied by $2$ because half the interval is defined for the integral): $$ \begin{eqnarray} V &=& 2 \pi \int_0^{b\sqrt{3}} [(2a)^2 - (\frac{a}{b}\sqrt{y^2+b^2})^2] \, \textrm{d}y \\ &=& \frac{2 \pi a^2}{b^2} \left[3 b^2 y - \frac{y^3}{3}\right]_0^{b \sqrt{3}} \\ &=& 4 \sqrt{3} \pi a^2 b \\ \end{eqnarray} $$

  2. Using volume of a solid of revolution with cylindrical-shell method: $$ \begin{eqnarray} V &=& 2 \pi \int_a^{2a} [x (\frac{2b}{a} \sqrt{x^2 - a^2})] \, \textrm{d}x \\ &=& 4 \pi a b \left[ \frac{(x^2 - a^2)^{3/2}}{3 a^3} \right]_a^{2a} \\ &=& 4 \sqrt{3} \pi a b \\ \end{eqnarray} $$

The second answer is the one from my book but I'd really like to understand why the first answer is different.

$\endgroup$
1
$\begingroup$

The second integral is wrong. \begin{equation*} \int \left(x\left(\frac{2b}{a}\sqrt{x^2-a^2}\right)\right)\,dx = \frac{2b}{3a}\int 3x\sqrt{x^2-a^2}\,dx = \frac{2b}{3a}(x^2-a^2)^{3/2}, \end{equation*} so the definite integral is \begin{equation*} V = 2\pi\int_a^{2a} \left(x\left(\frac{2b}{a}\sqrt{x^2-a^2}\right)\right)\,dx = \frac{4b\pi}{3a}(x^2-a^2)^{3/2}\big\lvert_a^{2a} = 4 \sqrt{3} \pi a^2 b . \end{equation*}

$\endgroup$
3
  • $\begingroup$ Thanks, so my book was wrong once again. By the way, what technique did you use to integrate $x \sqrt{x^2 - a^2}$? $\endgroup$
    – totembowl
    Jan 11 '16 at 3:08
  • $\begingroup$ Just substitution, essentially, after recognizing that $\frac{d}{dx}(x^2-a^2)^{3/2} = 3x$. I suppose the "calculus book" way of doing it is to use the substitution $x = a\sec u$ and the identity $\sec^2 u - 1 = \tan^2 u$. What book is this? $\endgroup$
    – rogerl
    Jan 11 '16 at 3:12
  • $\begingroup$ Thanks again, $x = a \sec \theta$ worked. My book is an old edition of The Calculus with Analytic Geometry by Louis Leithold. $\endgroup$
    – totembowl
    Jan 11 '16 at 3:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.