# Area enclosed by curve $f(x,y)$ defined implicitly

Find area bounded by the curve $$5x^2+2y^2+6xy+7x+6y+6=0$$

I can find the area using integration for curves defined explicitly in $$x$$ or $$y$$. I have no idea how to do this.

## 6 Answers

The hint.

By the affine transformation with determinant $$\Delta$$ write the equation of this ellipse in the form $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and the needed area is equal to $$\frac{\pi ab}{\Delta}$$

I got the following. $$\left(2x+y+\frac{1}{2}\right)^2+\left(x+y+\frac{5}{2}\right)^2=\frac{1}{2}.$$ Can you end it now?

• What is "affine transformation"? – tatan Jul 17 at 14:49
• @tatan see here: en.wikipedia.org/wiki/Affine_transformation It's enough to use rotation and translation. In this case $\Delta=1$. – Michael Rozenberg Jul 17 at 14:52
• @tatan Your equation is an ellipse but it's rotated and in an awkward place. An affine transformation rotates it and moves its center to $(0,0)$ which makes it easier to work with. – Jam Jul 17 at 14:59
• Yes, it is an ellipse. Can someone, please, tell me why Desmos graphs such an ellipse but with an unbounded branch located in the second and third quadrants? – Piquito Jul 17 at 15:30
• @Piquito I think you made some mistake. – Michael Rozenberg Jul 17 at 15:32

There is no simple general formula for such problems. In the case at hand it is expected that you recognize the curve as an ellipse $$E$$ whose axes are not parallel to the $$x$$- and the $$y$$-axes. Solving the given equation for $$y$$ should therefore result in two functions $$x\mapsto y_+(x)$$ and $$x\mapsto y_-(x)$$ describing the "upper half" and the "lower half" of the ellipse over some $$x$$-interval $$a\leq x\leq b$$. My computations gave $$y={1\over4}\left(-(6x+6)\pm\sqrt{4-(2x-4)^2}\right)\ .$$ This enforces $$|2x-4|\leq 2$$, or $$1\leq x\leq3$$. We therefore can say that $${\rm area}(E)=\int_1^3\bigl(y_+(x)-y_-(x)\bigr)\>dx={1\over2}\int_1^3 \sqrt{4-(2x-4)^2}\>dx={\pi\over2}\ .$$

• By my way we obtain: $\frac{\pi\cdot\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}}{1}=\frac{\pi}{2}.$ – Michael Rozenberg Jul 17 at 15:36

There's another way you can do it, just using basic calculus, if you're not comfortable with transformations. But it's tedious so @Michael_Rozenberg's answer is the best tactic.

Your equation is a polynomial with terms in $$y^2$$ and $$y$$ so you can think of it as a quadratic. Solve for $$y$$ using a method of your choice and rearrange, to get $$y=\frac{-\left(6x+6\right)\pm2\sqrt{-x^2+4x-3}}{4}$$. You can simplify the radical by factoring it into two integers, $$m, n$$. We can think of this as two functions: $$f^+(x)$$ and $$f^-(x)$$ when we take either the positive or negative value of the square-root. Each function corresponds to either an upper or lower curve since for each $$x$$ there were two roots to choose from. So, to take the area between the curves, we can find the difference between their integrals: $$A=\int f^+(x)\ \mathrm{d}x-\int f^-(x)\ \mathrm{d}x$$. We want to subtract them this way round since $$f^+(x)$$ is the upper curve.

We've found a closed form but where are we integrating over? Well, that's the interval where $$-x^2+4x-3>0$$ since otherwise the square-root would be undefined. So, we're integrating between those two integers you found earlier. Simplify the integral for $$A$$ with some elementary algebra and you should end up with $$A=\int_n^m\sqrt{(m-x)(x-n)}\ \mathrm{d}x$$. Integrate by substitution with $$u=x-2$$ and you've got $$\int_{-1}^1\sqrt{1-u^2}\ \mathrm{d}u$$. If you can remember the name of the curve $$y=\sqrt{1-x^2}$$, you're done.

All of the simpler methods that I can think of require first recognizing that the curve is an ellipse. One such method involves finding a parameterization $$\mathbf r(t)$$ relative to the center of the ellipse and then evaluating $$\frac12\int_0^{2\pi} \mathbf r(t)\wedge\mathbf r'(t)\,dt$$.

Let $$f(x,y)=5x^2+2y^2+6xy+7x+6y+6$$. The center of the ellipse can be found by solving $$\nabla f(x,y)=0$$, which produces $$\mathbf c=(2,-9/2)$$. Similarly, we can find a point where the tangent to the ellipse is vertical by solving $$f(x,y)=f_x(x,y)=0$$. The point $$\mathbf p_1=(1,-3)$$ looks convenient. To complete the parameterization, we need a conjugate diameter to $$\mathbf c\mathbf p_1$$, but since the tangent at $$\mathbf p_1$$ is vertical, this is just a matter of finding a point on the ellipse at $$x=2$$; $$\mathbf p_2=(2,-5)$$ will do. A parameterization of the ellipse is therefore $$\mathbf c+(\mathbf p_1-\mathbf c)\cos t+(\mathbf p_2-\mathbf c)\sin t.$$ For computing the area, we can translate the center to the origin by dropping $$\mathbf c$$, leaving $$\mathbf r(t) = \left(-\cos t,-\frac12(\cos t+3\sin t)\right).$$ We then have \begin{align} \mathbf r(t)\wedge\mathbf r'(t) &= \begin{vmatrix}-\cos t & -\frac12(\cos t+3\sin t) \\ \sin t & -\frac12(3\cos t-\sin t)\end{vmatrix} \\ &= \frac12 (3\cos^2 t - \cos t\sin t + \cos t\sin t +3\sin^2 t) \\ &= \frac12 \end{align} giving for the area $$\frac14\int_0^{2\pi}dt = \frac\pi2$$.

With a pair of conjugate diameters in hand, the ellipse’s area can also be computed directly: its area is equal to $$\pi a b$$, where $$a$$ and $$b$$ are the semiaxis lengths, but the area of the triangle with sides given by conjugate half-diameters of an ellipse is constant, so the ellipse’s area is equal to the absolute value of $$\pi \begin{vmatrix}\mathbf c & 1 \\ \mathbf p_1&1 \\ \mathbf p_2 & 1\end{vmatrix} = \frac\pi2.$$

The XY term can be eliminated by rotating the coordinate axis，Then you can judge the graph and calculate it by definite integral

just like Michael Rozenberg said, you can use the completing the square approach to make the equation take the form $$a^{2}x^{2}+b^{2}y^{2}=a^{2}b^{2}$$ the the result follows.