Question

Find the Volume of revolution of xy=x^3+3 between x=1 and x=2 about x axis

Answer 1

Well, I'd begin by rewriting the function in a more intuitive way. Divide both sides by x to get
\(\Large y = \frac{x^3}{x} + \frac{3}{x}\)
so
\(\Large y = x^2 +\frac{3}{x}\)

Answer 2

I agree, i got to that bit integrated and got 189/10 pi but answ is 179/10 pi not sure where i went wrong.

Answer 3

Should have been \[\int\limits_{1}^{2}\pi (x^2 +\frac{3}{x})^2dx\]

Answer 4

got that step

Answer 5

\[=\pi*\int\limits_{1}^{2}x^4 +3x + 3x +\frac{9}{x^2}dx\]

Answer 6

got that as well

Answer 7

\[=\pi \int\limits_{1}^{2}x^4 + 6x +9x^{-2}dx\]
I'd rewrite it that way, so that I could use exponent rule.

Answer 8

yes true after that u'd get pi [x^5/5 + 3x^2 -9/x] limits 2 and 1

Answer 9

\(\Large \pi*(\frac{x^5}{5} +3x^2 -\frac{9}{x})|_{1}^{2}\)

Answer 10

got it, fell at the last hurdle sub 1 into x^5/5 answ 0.2 not 1

Answer 11

Ah =) Glad you found your mistake, and I'm glad it was such a small one. Seems like you've got a good handle on the main concepts here and know how to do this =)

Answer 12

Not really took me a while to get to it. In the exam i wont have that time. Thanks for helping me thru it.

Answer 13

I certainly relate to that, buddy. As an undergrad, I was constantly running out of time on exams =/