x=y2 + 4y and x=0
someone help plz
This is a graph of a parabola turned on its right side. It intersects the y axis at y=0 & y=-4 (find this by solving for y in: $\displaystyle y^2+4y = 0$)
The area you are concerned with is the area of the parabola "cut off" by the y-axis (aka x=0).
Since you are dealing with functions in terms of y, your integral will be:
$\displaystyle \int_{c}^{d} (h(y) - k(y)) dy$
Where c & d are your limits of integration (on the y axis), h(y) is the "most positive" function and k(y) is the less positive function. By "positive" I mean the function that is furthest to the the right side of the graph
Can you try it from here?