Can you grunt it out

[BigGlenntheHeavy]

$\displaystyle Find \ volume \ of \ the \ solid \ formed \ by \ revolving \ the \ region \ bounded \ by \ the \ graphs \ of$

$\displaystyle y \ = \ x^{3}+x+1, \ y \ = \ 1, \ and \ x \ = \ 1 \ about \ the \ line \ x \ = \ 2, \ using \ the \ disc \ method, \ not \ shells.$

$\displaystyle Note, \ you'll \ have \ to \ solve \ y \ = \ x^{3}+x+1 \ for \ x.$

 

[BigGlenntheHeavy]

$\displaystyle OK, \ f(x) \ = \ y \ = \ x^{3}+x+1, \ hence \ f(x) \ has \ one \ real \ negative \ solution \ and \ two \ imaginary \ ones.$

$\displaystyle For \ f(x), \ we \ get \ x= \ \bigg(\frac{y-1}{2}+\sqrt\frac{4+27(1-y)^{2}}{108}\bigg)^{1/3}+\bigg(\frac{y-1}{2}-\sqrt\frac{4+27(1-y)^{2}}{108}\bigg)^{1/3}$

$\displaystyle [Washer \ Method]$

$\displaystyle V=\pi\int_{1}^{3}\bigg[\bigg(\bigg[\bigg(\bigg(\frac{y-1}{2}+\sqrt\frac{4+27(1-y)^{2}}{108}\bigg)^{1/3}+\bigg(\frac{y-1}{2}-\sqrt\frac{4+27(1-y)^{2}}{108}\bigg)^{1/3}\bigg)-2\bigg]^{2}\bigg)-1\bigg]dy=$

$\displaystyle \frac{29\pi}{15}$

$\displaystyle [Shell \ Method]$

$\displaystyle Definitely \ Shell \ Method$

$\displaystyle V=2\pi\int_{0}^{1}(2x-x^{2}+2x^{3}-x^{4})dx \ = \ \frac{29\pi}{15}$

$\displaystyle For \ further \ elucidation, \ see \ graph \ below.$

[attachment=0:3s7482px]abc.jpg[/attachment:3s7482px]

 

[galactus]

Cool, Glenn. Is that Cardano's method for solving a cubic?. I have not seen, nor used, that in years

 

[BigGlenntheHeavy]

It was one of the pre-renaissance Italians, I think there were two - Cardano was one of them.

With today's technology, there isn't any need to solve cubics algebraically, as one's trusty TI-89 will give you all and any solution(s); just killing time in my "Golden Years".

 

[BigGlenntheHeavy]

galactus, I'm still waiting for my new toy, should get it any day now.

 

[galactus]

Yeah, that's right. Let me know when it arrives.