Question

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.

Answer 1

x = -(1/6)y^2 + 13/6 ; x = 2y

Answer 2

id integrate with respect to y in the end; maybe :)

Answer 3

when you integrate with respect to y, do you have the it in terms of x or y ? im a little confused D:

Answer 4

terms of y

Answer 5

we need to equate the 2 equations i wrote and find the 'boundarys'

Answer 6

(-1/6)y^2 -2y +13/6 = 0 will give us the interval between the ys

Answer 7

multiply by -6 to normalize this thing right?

y^2 +12y -13 = 0

Answer 8

i think it should be integrated wrt x.check the curve please...which i have given...

Answer 9

y = -13, and y = 1

Answer 10

coulda tried to save it as a jpeg ya know :) loads faster

Answer 11

wrt x is fine; but trickier; wrt y is easier and more straight forward

Answer 12

for the region left where x=2y intersects the parabola. the y is x/2, and for the right the y=sqrt(13-6x).

Answer 13

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?

Answer 14

so the integral would look something like this..
\[\int\limits_{-13}^{1} (-1/6)y^2 +13/6-2y)dy\]

Answer 15

i believe so :)

Answer 16

than you both so much :)

Answer 17

thank*

Answer 18

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

Answer 19

gotcha. thanks a bunch :)

Answer 20

youre welcome :)

Answer 21

it all worked out. :D you rock :)

Answer 22

:) thnx