It's a jelly bean!

Here is my stupid non-rigorous calculus approach. Consider a red vertical strip of area $dA$ The length of this strip $D$ can be found by solving for $y$ . This is just the euclidean distance between points obtained by intersecting a vertical line at $x$ $(x,\sqrt{1-x^2} - x) , (x,-\sqrt{1-x^2} - x )$ thus the distance $D = 2\sqrt{1-x^2}$ . We know that the small area is just the the length times width of the strip $dA = Ddx$ $dA = 2\sqrt{1-x^2} dx$ $A =2\int_{-1}^{1} {\sqrt{1-x^2} dx = \pi}$

All we need to do is to rotate the co-ordinate axes so that they coincide with the major and minor axes of the ellipse. This will render the ellipse in the standard form. But the beauty is that we don't even need to do all the calculations explicitly if we employ a tiny bit of linear algebra. First, notice that $x^2+(y+x)^2=2x^2+2xy+y^2=\vec{z}^{T}A\vec{z}$ where $A=\begin{pmatrix} 2 &1 \\ 1 &1 \end{pmatrix}$ and $\vec{z}=\begin{pmatrix}x \\ y \end{pmatrix}$ . Now let us write the eigen decomposition of $A$ as $A=U\Lambda U^{T}$ where $U$ is an orthonormal (rotation) matrix and $\Lambda$ is a diagonal matrix consisting of eigenvalues of $A$ . Since rotation does not change the area, we change the co-ordinates as follows $\vec{\zeta}=U^T\vec{z}$ . With this transformation, the equation of the ellipse becomes $\vec{\zeta}^{T}\Lambda\vec{\zeta}=1$ However, since $\Lambda$ is diagonal, the above is equivalent to $\lambda_1\zeta_1^2+\lambda_2\zeta_2^2=1$ This is an ellipse in standard form and its area $S$ is well-known to be $S=\frac{\pi}{\sqrt{\lambda_1\lambda_2}}=\frac{\pi}{\sqrt{\text{det}(A)}}$ Thus all we need to do is to compute $\text{det}(A)$ , which is $1$ in this case and the answer follows.

Use only the geometry way to solve the problem with no trig and calculus:

Step 1: $\begin{cases} x^2+(y+x)^2=1 \\ x^2+y^2=1 \end{cases}$

Subtract the equations we have $x(2y+x) = 0.$

First two solutions are obvious: $A=(0,1)$ and $B=(0,-1)$ . The other two solutions are $C=(\frac{2}{\sqrt 5}, -\frac{1}{\sqrt 5} )$ , and $D=(-\frac{2}{\sqrt 5}, \frac{1}{\sqrt 5} )$ .

Slope of line AD is $m = -\dfrac{1+\sqrt 5}{2}$ .

The major axis of the ellipse (a) can be calculated from $\begin{cases} x^2+(y+x)^2=1 \\ y = -\dfrac{1+\sqrt 5}{2}x \end{cases}$

$a = \dfrac{\sqrt 5 +1}{2}.$

The minor axis of the ellipse (b) can be calcuated from $\begin{cases} x^2+(y+x)^2=1 \\ y = \dfrac{2}{1+\sqrt 5}x \end{cases}$

$b = \dfrac{\sqrt 5 -1}{2}.$

Therefore, the area of the ellipse is

$Area = \pi ab = \pi \dfrac{\sqrt 5 +1}{2} \cdot \dfrac{\sqrt 5 -1}{2} = \boxed{\pi}.$

Let $x=r \cos \theta , y=r\sin \theta$ Substitute this into the equation to get, $r^2 (1+\cos^2 \theta + 2 \sin \theta \cos \theta ) = 1$ Now, we can put any value of $\theta$ and get $r$

So, first we minimize $(1+\cos^2 \theta + 2 \sin \theta \cos \theta )$ to get semi-major axis by differentiating and equating to zero.

Similarly,we find semi-minor axis and the use the formula $A= \pi a b$

Let $u=x$ , $v=y+x$ . The Jacobian of the transformation from the $xy$ -plane to the $uv$ -plane is $\bigg|\text{det}\begin{pmatrix}1&1\\0&1\end{pmatrix}\bigg|=1$ , so area is preserved. In the $uv$ -plane, the ellipse is mapped to the unit circle given by $u^2+v^2=1$ . Therefore, the area of the ellipse is $\pi\approx\boxed{3.14}$ .

The ellipse implicitly defined by the equation $x^2 + (x+y)^2 = 1$ can be described by the following parametric funtion $\gamma(t) = ( cos(t), sin(t) - cos(t) )$ with $t \in [0, 2\pi]$ . On other hand let $F(x,y) = (0,x)$ be a vector field such as $\frac{\partial F_y}{\partial x}- \frac{\partial F_x}{\partial y} = 1$ so we can use the Green's Theorem:

$\int_0^{2\pi} F(\gamma(t)) \cdot \gamma'(t) dt = \int_\gamma\frac{\partial F_y}{\partial x}- \frac{\partial F_x}{\partial y} dA$

Replacing what we know in the above equation :

$\int_0^{2\pi} (0, cos(t)) \cdot (-sin(t), cos(t)-sin(t)) dt = \int_\gamma dA$

$\int_0^{2\pi} cos^2(t)-cos(t)sin(t) dt = A$

We know that $\int_0^{2\pi} cos^2(t) dt = \pi$ and $\int_0^{2\pi} -cos(t)sin(t) dt = 0$ then:

$A = \pi$

I resorted to Lagrange multipliers—something I can do symbolically for the most part.

Just use Cavalieri's principle

Let x=sinA

x+y=cosA or y=cosA-sinA

Integrate ydx from 0 to 2(pi) to get solution as pi

There are many ways to solve the problem. I can think of at least 4. 2 of which require calculus, one of which uses algebra+geometry (I wouldn't go as far as to call it analytical geom), and one which uses purely coordinate geometry.

$y=-x\pm\sqrt{-x^2+1}$ Area $=2 \int_{-1}^1 \sqrt{1-x^2}dx=\pi (1)^2$

i just found the inersection by the line y=-x nd the r s we ll know is pi l b , l is mjor xis nd b is minor xis , you get pi .

I believe this solution is simpler...

Note that the graph is a circle, except each vertical slice is just translated up/down by some amount, determined by x. However, simply moving the slices shouldn't change the area. Thus, the area should still be that of a unit circle, or $\pi$ .