# Finding average value

1. Nov 20, 2009

### tnutty

Find the average value of y on the region D that is bounded above by the parabola y = 4 - x2 and below by the x-axis.

Does this question makes sense?

I know the formula for the average values function :

1/(A(x)) * doubleIntegralOf F(x,y)DA

But its says above the parabola and below the x-axis, does that contradict each other?

There is no explanation in the answer but here it is :

The area of the region is 64/3 and the double integral of y over the region is 256/15. So
y_avg = 4/5.

Can you explain and show me how to do this ?

2. Nov 20, 2009

### clamtrox

"bounded above" means that the parabola is the upper limit.

3. Nov 20, 2009

### tnutty

so is it just this :

1/2$$\int^{2}_{0} 4 - x^2$$

And how about the double integral of it?

4. Nov 20, 2009

### HallsofIvy

Staff Emeritus
The parabola, y= 4- x2, intersects the x-axis at (-2, 0) and (2, 0). On the entire region, bounded by the parabola and the x-axis, x ranges from -2 to 2. For each x, y ranges between the x-axis and the parabola- from y= 0 to y= 4- x2. Those are the limits of integration:
$$\int_{x=-2}^2 \int_{y= 0}^{4- x^2} y dydx$$
Of course, because of the symmetry, you can integrate from x=0 to x= 2 and double.

Notice that the double integral for the area is just
$$\int_{x=-2}^2 \int_{y= 0}^{4- x^2} dydx[/itex] and becomes [tex]\int_{x=-2}^2 \left[y\right]_{y=0}^{4-x^2}= \int_{x=-2}^2 (4- x^2)dx$$
$$= 2\int_0^2 4- x^2 dx$$