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

Question

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.

Answer 1

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?

Answer 2

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.

Answer 3

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 $${\rm area}(E)=\int_1^3\bigl(y_+(x)-y_-(x)\bigr)\>dx={1\over2}\int_1^3 \sqrt{4-(2x-4)^2}\>dx={\pi\over2}\ .$$

Answer 4

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.

Answer 5

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 $$\pi \begin{vmatrix}\mathbf c & 1 \\ \mathbf p_1&1 \\ \mathbf p_2 & 1\end{vmatrix} = \frac\pi2.$$

Answer 6

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