find the area of the region bounded by the curves x+y=0 and x=y^2+3y?

Question

would someone please help me with this problem? im not too sure how to solve it..

Answer 1

The first step in solving this is to determine the bounds of integration. We need to know where these two curves intersect.

First off, note that x + y = 0 means x = -y. But x = y^2 + 3y, so we equate these two.

-y = y^2 + 3y

And bringing the -y over to the right hand side gives us

0 = y^2 + 4y, which we can factor as
0 = y(y+4), therefore
y = {0, -4}

Those are our bounds of integration. Note that since we're working with y, we're solving our area problem vertically.

Whenever we deal with areas between curves in terms of x, what we'd normally do is solve it in this fashion:

A = Integral (a to b, "higher curve" minus "lower curve" dx)

Or to be more mathematically formal, for curves f(x) and g(x), if
f(x) is greater than g(x) on the interval a to b, then our formula would be

A = Integral (a to b, [f(x) - g(x)] dx)

In our case though, since we're working with y, we want to determine which curve is "righter" or "more to the right". The best way to determine which curve is "righter" is to test a value between our bounds -4 and 0. Let's test -1.

When y = -1, which curve is greater: x = -y, or x = y^2 + 3y?
Testing, we get x = -(-1) = 1 for the first curve, and x = 1 - 3 = -2 for the second curve. Clearly, the first curve is greater, so that's what we write first.

A = Integral (-4 to 0, [-y] - [y^2 + 3y] )dy
A = Integral (-4 to 0, -y - y^2 - 3y) dy
A = Integral (-4 to 0, -4y - y^2)dy

And now we solve the integral.

A = [-2y^2 - (1/3)y^3] {evaluated from -4 to 0}, so
A = [0 - 0] - [-2 (-4)^2 - (1/3)(-4)^3 ]
A = - [-32 - (-64/3)]
A = - [-32 + 64/3]
A = - [-96/3 + 64/3]
A = - [-32/3] = 32/3

Answer 2

First find cross-points: -y=y^2+3y or y(y+4)=0, hence y1=0, y2=-4;
Elementary area ds=L*dy inside the region, that is L=(-y) - (y^2+3y) or (straight line x=-y) minus (parabola x=y^2+3y).
Now S=integral{for y=-4 until 0} of (-y^2-4)*dy = -y^3/3-4y = 0 - (-64/3+16) = 16/3;

Answer 3

I'll tell you the steps to do this problem:

1. Find the points of intersection of the two curves. You will find there are two such points. Take the x-coordinates of those points as "a" and "b".

2. Integrate the first curve w.r.t "x" from x=a to x=b. Take is value to be INT1

3. Integrate the second curve w.r.t "x" from x=a to x=b. Take is value to be INT2

4. Subtract INT1 from INT2. This will give you the answer.

Answer 4

hey..i had a similar question..check out http://www.schoolpiggyback.com ...its expert students that do the assignment for you...its worth a shot..goodluck : )

Answer 5

x + y = 0 is a straight line, which forms is -45o with the x axis and passes through the origin of coordinates (0,0)

y = x^2 + 3x is a parabola whose roots are 0 and - 3. Draw it and consider that x = y^2 + 3y is the same kind of parabola, but its axis will be horizontal.

So, you will have this:
.....................................................o
...................o
........o
....o
.o
o______________________________>x
.o..x
....o.....x
.........o......x
.....................xo
.........................x.............................o
..............................x

Ana