SOLUTION: Let R be the region bounded by the curves y =root(x + 1 )and y = (x + 1)^2
. We will find the
volume of the solid obtained by rotating R about the horizontal line y = −1.
Algebra.Com
Question 1117974: Let R be the region bounded by the curves y =root(x + 1 )and y = (x + 1)^2
. We will find the
volume of the solid obtained by rotating R about the horizontal line y = −1.
(a) Find the points of intersection of the two curves.
(b) Sketch a graph of a portion of the curves, the region R, and the line y = −1.
(c) Calculate the volume of the solid obtained by rotating R about the horizontal line y = −1.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Points of Intersection
^2 =\ \sqrt{x\ +\ 1})
^4 =\ x\ +\ 1\)
\ =\ 0)
(x^2\ +\ 3x\ +\ 3)\ =\ 0)
Two real roots, 
and 
\ =\ \sqrt{0\ +\ 1}\ =\ 1)
\ =\ \sqrt{-1\ +\ 1}\ =\ 0)
Points of intersection are \ \ )
and )
Sketch
Volume of 
rotated around 
^2\ -\ \((x\ +\ 1)^2\ +\ 1\)^2\,dx)
^{\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
My calculator said it, I believe it, that settles it
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