# Extra Points on Test!

#1

Show work and answers for the volume of the solid created by rotating the area between:
y = x^2 and y = 2 -x around:

a) the line y = -1
b) the y-axis and
c) the line x = -2

#2

a. around the line y= -1 using the Washer Method

  Finding the intersection points-


x^2=2-x
x^2+x-2=0
(x+2)(x-1)=0 \space \rightarrow \space x = -2 , x =1

y= x^2 is the bottom function “r”
y= 2-x is the top function “R”

Use equation: \pi \int (R - axis \space of \space rotation)^2 - (r - axis \space of \space rotation)^2

\pi \int_{-2}^1 (2-x+1)^2 - (x^2+1)^2 dx

simplifies to
\pi \int_ {-2}^1 (3-x)^2 - (x^2+1)^2 dx
then square both…
\pi \int_ {-2}^1 (x^2-6x+9) - (x^4+2x^2+1) dx
then combine like terms…
\pi \int_ {-2}^1 (-x^4-x^2-6x+8) dx
Integrate…
\Pi [(-1/5x^5-1/3x^3-3x^2+8x)] between [1, -2]

\pi [( -1/5(1)^5 - 1/3(1)^3 - 3(1)^2 + 8 (1)) - ( -1/5(-2)^5 - 1/3(-2)^3 - 3(-2)^2 + 8 (-2))]
= \pi [(4.466666667)+(18.93333333)]
= 73.5132681 units^2

b. around the line x= -2 , using the Shell Method Equation: 2\pi \int r * h * d-something
r= x-axis of rotation h = ( top - bottom)
Finding the intersection points-
x^2=2-x
x^2+x-2=0
(x+2)(x-1) x = -2 , x =1

r= (x+2) h= (2-x-x^2) d-something = dx

2\pi\int_{-2}^1 (x+2) (2-x-x^2) dx
foil…
2\pi\int_{-2}^1 (2x - x^2 - x^3 + 4 - 2x - 2x^2) dx
combine like terms…
2\pi\int_{-2}^1 ( -x^3 - 3x^2 +4) dx
Integrate
2\pi[ -1/4x^4 - x^3 +4x ] from_{-2}^1

2\pi [( -1/4(1)^4 - (1)^3 +4(1)) - (-1/4(-2)^4 - (-2)^3 +4(-2))
= 2\pi [ (2.75) - (-4) ]
= 2\pi(6.75)
=13.5 \pi
= 42.41150082 units^2

c. around the line y-axis using the Shell Method.
2\pi\int_{-2}^1 r * h * d-something
r= x-axis of rotation h = ( top - bottom)

2\pi\int_{-2}^0 (x+0) (2-x-x^2) dx
foil
2\pi\int_{-2}^0 (2x-x^2-x^3) dx
integrate
2\pi (x^2 - 1/3x^3 - 1/4x^4) between [1, -2]
2\pi[ (0^2 - 1/3 0^3 - 1/4 0^4) - ( -2 ^2 - 1/3 (-2)^3 - 1/4 (-2)^4) ]
2\pi [ 0 - (8/3)]
= absolute value of -16.75516082 units^2

#3

Jennifer, I have not checked the math, but look at the edits that were made to make it look “pretty”. See if you can finish it. The main thing is to use \$ symbols around the math stuff.
A few codes: \int is the integral sign, \pi gives you pi

#4

First answer is right. Check the answer around the line x = -2. Set up is good, but I get a different answer. For around the y-axis, I realized that it goes back into itself. Thus the first piece from -2 to 0 is all you need.

#5

a) \displaystyle V= \pi \int\limits_{-2}^1 (2-x+1)^2-(x^2+1)^2 dx = \pi \int\limits_{-2}^1 (3-x)(3-x)-(x^2+1)(x^2+1)dx
\displaystyle = \pi \int\limits_{-2}^1 9-6x+x^2-x^4-2x^2-1dx = \pi \int\limits_{-2}^1 -x^4-x^2-6x+8dx
\displaystyle = \pi [\dfrac{-x^5}{5}-\dfrac{x^3}{3}-3x^2+8x] \Big|_{-2}^1 = \pi [(\tfrac{-1}{5}-\tfrac{1}{3}-3+8-(\tfrac{32}{5}+\tfrac{8}{3}-12-16)]
\displaystyle = \pi[\tfrac{-33}{5}-\tfrac{9}{3}+33] = \tfrac{117\pi}{5} units^2 when rotated about \displaystyle y=-1

b) Calculate negative portion of graph \displaystyle V= 2\pi \int\limits_{-2}^0x(2-x-x^2)dx = 2\pi \int\limits_{-2}^02x-x^2-x^3dx = 2\pi [\dfrac{2x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}]\Big|_{-2}^0
\displaystyle = 2\pi[0-(4+\tfrac{8}{3}-\tfrac{16}{4})] =- \dfrac{16\pi}{3} units

Calculate positive portion of graph \displaystyle V = 2\pi \int\limits_{0}^1x(2-x-x^2)dx = 2\pi \int\limits_{0}^12x-x^2-x^3dx = 2\pi [\dfrac{2x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}]\Big|_{0}^1
\displaystyle = 2\pi[1-\tfrac{1}{3}-\tfrac{1}{4}-0] = \dfrac{5\pi}{6} units

Add volumes of both portions together

\dfrac{5\pi}{6}+\dfrac{16\pi}{3} = \dfrac{36\pi}{6}units^2

c) \displaystyle V = 2\pi \int\limits_{-2}^1(x+2)(2-x-x^2)dx = 2\pi\int\limits_{-2}^1 2x-x^2-x^3+4-2x-2x^2dx
\displaystyle = 2\pi \int\limits_{-2}^1-x^3-3x^2+4dx = 2\pi[-\dfrac{x^4}{4}-x^3+4x] \Big|_{-2}^1 = 2\pi[-\tfrac{1}{2}-1+4-(-\tfrac{16}{4}+8-8)]
\displaystyle = 2\pi(-\tfrac{3}{2}+8) = 2\pi(\tfrac{13}{2}) = 13\pi units^2

#6