# How to interpret negative surface area?

In calculating $$\iint_Dx^2y-y^5 dxdy$$

where D is given by:$$~~~~~1-y^2\leq x\leq 2-y^2\\-\sqrt{1+x}\leq y\leq\sqrt{1+x}$$

I refered to the graphs in the following link: enter link description here to determine the region of integration and reverse the order of integration by switching domains to:$$0\leq x\leq2\\\sqrt{1-x}\leq y\leq\sqrt{2-x}\\$$

From this I found that $$\int_0^2\int_\sqrt{1-x}^\sqrt{2-x} x^2y-y^5dydx=\frac{2}3$$ Which is multiplied by 2 for the total area, since $$\frac{2}3$$ only covers the first quadrant

But then the strip that is above $$y=\sqrt{1+x}$$ comes along with it, so I figured I must calculate the following integral and subtract it from $$\frac{2}3$$:$$\int_0^{1/2}\int_\sqrt{1+x}^\sqrt{2-x}x^2y-y^5dydx=-\frac{9}{32}\\$$

Why is this negative? And how should I interpret it? I wanted to subtract but should I now add it? Both calculations are correct but the negative area makes me think that there must be some incorrect reasoning in integrating x from 0 to 1/2 and then subtracting it from the first area

• The negative value is caused by the integrand in which the $-y^5$ is big as it is greater than 1. Furthermore it is not an area integral so negative area is not a concern. Also your first integral should allow for x from 1 to 2 in which the y limits should be from 0 to $\sqrt{2 - x}$. – KY Tang Mar 18 at 0:56
• Thank you, but allowing for x from 1 to 2, do you mean this would make three integrals with x from 0to1/2, from 1/2to1 and 1to2 and then adding them up? – TheMercury79 Mar 18 at 5:10
• Yes please see my answer below. Thanks – KY Tang Mar 18 at 13:15

There should be 5 integrals: $$I_1 = \int_1^2\int_{-\sqrt{2 -x}}^\sqrt{2 - x}(x^2y - y^5)dy dx = 0$$
$$I_2 = (\int_{\frac{1}{2}}^{1}\int_{\sqrt{1 - x}}^{\sqrt{2 - x}} + \int_\frac{1}{2}^{1}\int_{-\sqrt{2 - x}}^{-\sqrt{1 - x}})(x^2y - y^5)dy dx = 0$$
$$I_3 = (\int_0^\frac{1}{2}\int_{\sqrt{1 - x}}^{\sqrt{1 + x}} + \int_0^{\frac{1}{2}}\int_{-\sqrt{1 + x}}^{-\sqrt{1 - x}})(x^2y - y^5)dy dx = 0$$
Your picture shows that the domain of integration is mirror symmetric with respect to the $$x$$-axis. Since the integrand is an odd function of $$y$$ it follows that the value of the integral is $$0$$.