- #1

CellCoree

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Sketch the region enclosed by [tex]x+y^2=12[/tex] and [tex]x+y=0[/tex]. Decide whether to integrate with respect to [tex]x[/tex] or [tex]y[/tex]. Then find the area of the region.

first thing i did was solve for x for each equation then set them to each other and got:

[tex]12-y^2=-y[/tex]

[tex]y^2-y-12=0[/tex]

found the points of intersection of y at 4,-3

plugged in for y and found the x intersections and got:

(-4,4) & (3,-3) for the points of intersections

ok time to set up the integral with respect to y

[tex]\int (-y) -(12-y^2)[/tex] (a=-3, b=4, don't know how to set those up using latex)

add the terms together to make it more neat...

[tex]\int y^2-y-12[/tex] (a=-3, b=4)

integrate...

[tex] \frac{y^3}{3} - \frac{y^2}{2} - 12y[/tex] (a=-3,b=4)

[tex] \frac{4^3}{3} - \frac{4^2}{2} - 12(4)[/tex]

subtract the above part by...

[tex] \frac{-3^3}{3} - \frac{-3^2}{2} - 12(-3)[/tex]

and got the answer -57.16667, which i know is way off. what am i doing wrong??

first thing i did was solve for x for each equation then set them to each other and got:

[tex]12-y^2=-y[/tex]

[tex]y^2-y-12=0[/tex]

found the points of intersection of y at 4,-3

plugged in for y and found the x intersections and got:

(-4,4) & (3,-3) for the points of intersections

ok time to set up the integral with respect to y

[tex]\int (-y) -(12-y^2)[/tex] (a=-3, b=4, don't know how to set those up using latex)

add the terms together to make it more neat...

[tex]\int y^2-y-12[/tex] (a=-3, b=4)

integrate...

[tex] \frac{y^3}{3} - \frac{y^2}{2} - 12y[/tex] (a=-3,b=4)

[tex] \frac{4^3}{3} - \frac{4^2}{2} - 12(4)[/tex]

subtract the above part by...

[tex] \frac{-3^3}{3} - \frac{-3^2}{2} - 12(-3)[/tex]

and got the answer -57.16667, which i know is way off. what am i doing wrong??

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