region enclosed

[CellCoree]

Sketch the region enclosed by x+y^2=12 and x+y=0. Decide whether to integrate with respect to x or y. 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:
12-y^2=-y
y^2-y-12=0

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

\int (-y) -(12-y^2) (a=-3, b=4, dont know how to set those up using latex)

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

\int y^2-y-12 (a=-3, b=4)

integrate...

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


\frac{4^3}{3} - \frac{4^2}{2} - 12(4)
subtract the above part by...
\frac{-3^3}{3} - \frac{-3^2}{2} - 12(-3)


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

[Pyrrhus]

At first glance, i'll say check your signs for the f(b) - f(a).

[CellCoree]

k checked that part and got the same signs, so what is wrong with my answer? did i set up the equation correctly? (is there a Ti-83 plus program i can use to check my answer? or is there any math program for the computer that would do this for me?

[diegojco]

well you can integrate in terms of x or try a change of variable, but its easy to get the area directly.

first you have the area enclosed between the straight line y=-x and the horizontal parabola y^2=-x+12. to find the points of intersection substitute y=-x in the parabola ecuation and you will get a second grade ecuation for x that haves the solutions -4 and 3, then in x=-4 and x=3 are the intersections.

second, make a sketch of the situation and think the area of the enclosed region between -4 and 0 is the integral of the parabola minus the integral of the line. and between 0 and 12 are 2 times the integral of the parabola minus a part of this (the two times is simply because with the integral of the parabola you only get the area above x axis and below the parabola and by symmetry the area between 0 and 12 is the same above x and below x).

third, the part that you must substract is the area enclosed between the y=x x=0 and the parabola that it's trivial to get.

then: where int(([a,b],f(x)) is integral of f(x) between a and b and sqrt square root:

A=int([-4,0],sqrt(-x+12)+x)+2*int([0,12],sqrt(-x+12))-int([0,3],sqrt(-x+12)-x)

A=(2/3)*(16^(3/2)+9^(3/2))-(5/2)=58,1666=58,2