x=y2 + 4y and x=0

someone help plz

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- Apr 14th 2009, 07:39 PMemoryfind area bounded by graph
x=y2 + 4y and x=0

someone help plz - Apr 14th 2009, 08:10 PMmollymcf2009
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?