How to find surface area of a wing by integration

Question

I have a question about aerodynamics. I came across that the wing area of ​​an airplane is calculated with this formula, but I can't calculate the integral. Even if the plane wing area is not calculated like this, can you please solve it step by step, it is very important to me. Wing planform geometry. Do not get hung up on the unnecessary details in the picture(or you could If you wish) $$\frac{\left(\int_{-\frac{b}{2}}^{\frac{b}{2}}c\ dy\right)}{\int_{-\frac{b}{2}}^{\frac{b}{2}}\ dy}b$$ Where c represents the chord length at a given distance y from the centerline. And $$b=11$$ $$c_{t}=1.63$$ $$c_{0}=1.12$$ unit

Answer 1

Because the wing-length isn't given, you don't know $c(y)$; all you can say from the figure is:

$$c(y)=\cases{\frac{b+2y}{b-w}c_0+\frac{-2y-w}{b-w}c_t & $\frac{-b}{2}\le y\le\frac{-w}{2}$ \\ c_0 & $-\frac{w}{2}<y\leq \frac{w}{2}$\\ \frac{b-2y}{b-w}c_0+\frac{2y-w}{b-w}c_t & $\frac{w}{2}<y\le\frac{b}{2}$} $$

where $w$ is your unknown fuselage width.

Decide on $w$ and the integral is routine.

Taking the integrals over only the wings (excluding the fuselage),

$$\begin{align}\frac{2\int_{\frac{w}{2}}^{\frac{b}{2}}\left(\frac{b-2y}{b-w}c_0+\frac{2y-w}{b-w}c_t\right)dy}{2\int_{\frac{w}{2}}^{\frac{b}{2}}dy}(b-w) &=\frac{\frac{\frac{b^2}{2}}{b-w}c_0+\frac{\frac{b^2}{2}-bw}{b-w}c_t}{b-w}(b-w)-\frac{\frac{bw-\frac{w^2}{2}}{b-w}c_0-\frac{\frac{w^2}{2}}{b-w}c_t}{b-w}(b-w)\\ &=\frac{\frac{b^2}{2}+\frac{w^2}{2}-bw}{b-w}(c_0+c_t)\\ &=\frac{b-w}{2}(c_0+c_t) \end{align}$$

Of course our wingshape was assumed trapezoidal, and so this formula could have been written without the integration