# Finding the area of the part of the plane given by the curves

I am supposed to find the area of the part of the plane given by the curves: $$y=x^{2},y=-x^{2}, -4x+4=y$$ I know that I have to divide it into two parts and so on.. but my question is.. how do I know which function to put first and second in the formula:

$$\int _{a}^{b}(f(x)-g(x))dx$$

My idea for the first integral is : $$\int _{0}^{-2+2\sqrt{2}}(x^{2}+4x-4)dx$$

Is that correct?

• No, it isn't. The first function is that whose graph lies above the second. In this first integral, the functions should be $x^2$ and $-x^2.$ The limits of the integral are right. Oct 20, 2019 at 12:54
• @user376343 okay so the result would be $$80\sqrt{2}-112$$ and the second integral would be $$\int _{-2+2\sqrt{2}}^{2}(-4x+4+x^{2})dx$$? Oct 20, 2019 at 13:08
• The second integral is correct. Oct 20, 2019 at 15:19

Hint:

You can split into the area above the $$x$$-axis, which involves only the upper parabola and the straight line and the area below the $$x$$-axis, which involves the lower parabola and the line.

For each area, you can determine the intersection $$A, B$$ of the line with the upper and lower parabola respectively, and its $$x$$-intercept $$C=(x_0,0)$$. If $$\xi_+$$ and $$\xi_-$$ are the abscissæ of $$A$$ and $$B$$, $$H,K$$ their projections on the $$x$$-axis, the upper and lower areas are (see figure below): $$\mathcal A_+=\int_0^{\xi_+}\!\!x^2\,\mathrm dx+\mathcal A(\text{triangle }AHC),\qquad\mathcal A_-=\int^0_{\xi_-}\!\!x^2\,\mathrm dx-\mathcal A(\text{triangle }BKC).$$

• Can you please check my answer , which I posted if it is correct? Oct 20, 2019 at 13:31
• No it isn't, because on the interval $[0,2\sqrt2-2)$, the straight line is above the parabola. Oct 20, 2019 at 13:38
• si on interval $$\left [ 0,-2+2\sqrt{2} \right ]$$ it is not the $$\int (x^{2}-(-x^{2}))dx$$? Oct 20, 2019 at 13:47
• It is, but that's not what you wrote, unless I've misread. Oct 20, 2019 at 13:50
• Sorry I made a mistake.. and the second integral is correct? Oct 20, 2019 at 13:54

Hints. First, find the intersection points of the lines. It's immediate to see that the two parabolas intersect in only one point ($$x=0$$), so that they bound no area between them. Thus what you want to do is solve each one of the equations $$\pm x^2=-4x+4.$$

Then you find out in which intervals each parabola is above or below the line. That's what will give you an idea which to subtract from which. That is, in each determined interval, find out whether $$f-g$$ is positive or negative, where $$f$$ represents one of the parabolas and $$g$$ represents the straight line.