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Finding area and volume of bounded region via integration

  1. Apr 7, 2014 #1
    Hi,

    I just need these solutions checked.
    Thank you in advance!

    Consider the region bounded by the following curves ##y=x-3, y=5-x, \text{and}\ y=3##:

    1.) set up an integral expression that would give the area of the region of y as a function of x:

    ##y = x-3 = 5-x##
    ##x + x - 3 - 5 = 0##
    ##2x-8=0##
    ##x = 4##
    ##\displaystyle \int_0^3 |\left(2x-8\right)|\ dx##

    2.) set up an integral expression that would give the area of the region of x as a function of y:

    ##\displaystyle \int_0^5 |\left(5-y\right)|##

    3.) compute the area of the region by evaluating one of the expressions in (1.) or (2.):

    ##\displaystyle \int_0^3 |\left(2x-8\right)|\ dx##
    ##\left(x-8\right)+\ \text{C}##
    ##\displaystyle = \lim_{x \to 0^{+}} ((x-8)x)=0##
    ##\displaystyle = \lim_{x \to 3^{-}} ((x-8)x)=-15##
    ##\displaystyle |-15-0| = 15##


    4.) set up an integral expression that would give the volume of the solid created by rotating the region about y = 1, using the disk/washer method:

    ##\displaystyle \pi \int_0^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##


    5.) find the volume of the solid by evaluating the integral/integrals in (4.):

    ##\displaystyle \pi \int_0^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##
    ##=\ -\dfrac{29}{6}\pi##
    ##\approx 15\ \text{cubic units}##

    6.) set up an integral expression that would give the volume of the solid created by rotating the region about y = -2, using the disk/washer method:

    ##\displaystyle \pi \int_{-2}^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##


    7.) compute the volume of the solid by evaluating the integral or integrals in (6.):

    ##\displaystyle \pi \int_{-2}^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##
    ##=\ -\dfrac{27}{2}\pi##
    ##\approx 42\ \text{cubic units}##
     
  2. jcsd
  3. Apr 7, 2014 #2

    LCKurtz

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    I would suggest you start over and begin by drawing a picture of the region. You are going to need two integrals using either a dx or dy variable. Also, what bounds the region on the left?
     
  4. Apr 7, 2014 #3
    You were correct in identifying x=4 as an intersection point, and y = 1 there; but the other two intersection points must occur where y = 3, so x is certainly NOT zero or 3 ! So there is no reason to integrate from 0 to 3 with respect to x.
     
  5. Apr 13, 2014 #4
    Thanks LCKurtz and az_lender!

    1. ##\displaystyle \int_2^4 \left(3-\left(5-x\right)\right)dx\,+\,\displaystyle \int_4^6\left(\left(3-\left(x-3\right)\right)\right)dx##
    ::
    ::

    2. ##\displaystyle \int_1^3 ((y+3-(5-y))dy\,=\,\int_1^3 (2y-2))dy##
    ::
    ::

    3. Using (1.) from above:
    ##\displaystyle \int_2^4 \left(3-\left(5-x\right)\right)dx\,+\,\displaystyle \int_4^6\left(\left(3-\left(x-3\right)\right)\right)dx##
    ##=\,\dfrac{x^2}{2}\,+\,C##
    ##\displaystyle\lim_{x\to 2^{+}} \left(\dfrac{x^2}{2}\right)\,=\,2##
    ##\displaystyle\lim_{x\to 4^{-}} \left(\dfrac{x^2}{2}\right)\,=\,8##

    ##\text{From the above}\,=\,8-2\,=\,6##
    ::
    ::

    4. ##\displaystyle \int_2^4 \pi ((2)^2-((3-(5-x)))^2 dx\,+\,\int_4^6 \pi ((2)^2\,-\,(3-(x-3)))^2 dx##
    ::
    ::

    5. From (4.) above:
    ##\dfrac{16\left(1-3\pi\right)\pi}{3}##
    ::
    ::

    6. ##\displaystyle \int_1^3 \pi \left(6-\left(5-y\right)\right)^2\,-\,\left(6\,-\,\left(y\,+\,3\right)\right)^2dy\
    \left(6-\left(5-y\right)\right)^2\,-\,\displaystyle \int \left(6\,-\,\left(y\,+\,3\right)\right)^2dy\
    \,=\,4\left(y\,-\,2\right)y\,+\,\dfrac{28}{3}\,+\,C##
    ::
    ::

    7. From (6.) above:
    ##\displaystyle\lim_{y\to 1^{+}} (4(y-2))y\,+\,\dfrac{28}{3})\,=\,\dfrac{16\pi}{3}##
    ##\displaystyle\lim_{y\to 3^{-}} (4(y-2))y\,+\,\dfrac{28}{3})\,=\,\dfrac{64\pi}{3}##

    ##=\,\dfrac{64\pi}{3}\,-\,\dfrac{16\pi}{3}\,=\,16\pi##
     
    Last edited: Apr 13, 2014
  6. Apr 13, 2014 #5

    LCKurtz

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    Those look correct for the triangular area.
     
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