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anonymous
 5 years ago
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.
anonymous
 5 years ago
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.

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x = (1/6)y^2 + 13/6 ; x = 2y

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0id integrate with respect to y in the end; maybe :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you integrate with respect to y, do you have the it in terms of x or y ? im a little confused D:

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we need to equate the 2 equations i wrote and find the 'boundarys'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(1/6)y^2 2y +13/6 = 0 will give us the interval between the ys

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0multiply by 6 to normalize this thing right? y^2 +12y 13 = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think it should be integrated wrt x.check the curve please...which i have given...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0coulda tried to save it as a jpeg ya know :) loads faster

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0wrt x is fine; but trickier; wrt y is easier and more straight forward

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the region left where x=2y intersects the parabola. the y is x/2, and for the right the y=sqrt(136x).

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when 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
 5 years ago
Best ResponseYou've already chosen the best response.0so the integral would look something like this.. \[\int\limits_{13}^{1} (1/6)y^2 +13/62y)dy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0than you both so much :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if it turns out negative; just take the absolute value; it simply means you subtracted in the wrong order for example: 53 = 2 35 = 2; but 2 = 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gotcha. thanks a bunch :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it all worked out. :D you rock :)
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