Question 1117974
<font face="Times New Roman" size="+2">


<b>Points of Intersection</b>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(x\ +\ 1)^2 =\ \sqrt{x\ +\ 1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 1)^4 =\ x\ +\ 1\]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4\ +\ 4x^3\ +\ 6x^2\ +\ 4x\ +\ 1\ =\ x\ +\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4\ +\ 4x^3\ +\ 6x^2\ +\ 3x\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x^3\ +\ 4x^2\ +\ 6x\ +\ 3)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x\ +\ 1)(x^2\ +\ 3x\ +\ 3)\ =\ 0]


Two real roots, *[tex \Large x\ =\ 0\ \ ] and *[tex \Large x\ =\ -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(0)\ =\ \sqrt{0\ +\ 1}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(-1)\ =\ \sqrt{-1\ +\ 1}\ =\ 0]


Points of intersection are *[tex \Large (0,1)\ \ ] and *[tex \Large (-1,0)]


<b>Sketch</b>


*[illustration Volume_of_Rotation.jpg]


<b>Volume of *[tex \Large R] rotated around *[tex \Large y\ =\ -1]</b>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V\ =\ \pi\int_{-1}^0\,\(\sqrt{x\ +\ 1}\ +\ 1\)^2\ -\ \((x\ +\ 1)^2\ +\ 1\)^2\,dx]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ \frac{\pi\(40\(x\ +\ 1\)^{\frac{3}{2}}\ -\ 6x^5\ -\ 30x4\ -\ 80x^3\ -\ 105x^2\ -\ 60x\)}{30}\|_{-1}^0\ =\ \frac{29\pi}{30}]


Just slightly north of 3 cubic units.  Verification of the antiderivative is left as an exercise for the student.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>