anonymous
  • anonymous
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. 6x+y^2=13, x=2y find area of the region.
Mathematics
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anonymous
  • anonymous
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. 6x+y^2=13, x=2y find area of the region.
Mathematics
chestercat
  • chestercat
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amistre64
  • amistre64
x = -(1/6)y^2 + 13/6 ; x = 2y
amistre64
  • amistre64
id integrate with respect to y in the end; maybe :)
anonymous
  • anonymous
when you integrate with respect to y, do you have the it in terms of x or y ? im a little confused D:

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amistre64
  • amistre64
terms of y
amistre64
  • amistre64
we need to equate the 2 equations i wrote and find the 'boundarys'
amistre64
  • amistre64
(-1/6)y^2 -2y +13/6 = 0 will give us the interval between the ys
anonymous
  • anonymous
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amistre64
  • amistre64
multiply by -6 to normalize this thing right? y^2 +12y -13 = 0
anonymous
  • anonymous
i think it should be integrated wrt x.check the curve please...which i have given...
amistre64
  • amistre64
y = -13, and y = 1
amistre64
  • amistre64
coulda tried to save it as a jpeg ya know :) loads faster
amistre64
  • amistre64
wrt x is fine; but trickier; wrt y is easier and more straight forward
anonymous
  • anonymous
for the region left where x=2y intersects the parabola. the y is x/2, and for the right the y=sqrt(13-6x).
amistre64
  • amistre64
when we integrate this area, we need to subtract one function from the other to get the total area; do we do absolute values? or let the areas lie where they are either pos or neg?
anonymous
  • anonymous
so the integral would look something like this.. \[\int\limits_{-13}^{1} (-1/6)y^2 +13/6-2y)dy\]
amistre64
  • amistre64
i believe so :)
anonymous
  • anonymous
than you both so much :)
anonymous
  • anonymous
thank*
amistre64
  • amistre64
if it turns out negative; just take the absolute value; it simply means you subtracted in the wrong order for example: 5-3 = 2 3-5 = -2; but |-2| = 2
anonymous
  • anonymous
gotcha. thanks a bunch :)
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
it all worked out. :D you rock :)
amistre64
  • amistre64
:) thnx

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