What is area x=3y^2,x=4+2y^2 ?

The answer to area x=3y^2,x=4+2y^2 is No Solution

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area x=3y^2,x=4+2y^2

area

x = 3 y^{2}, x = 4 + 2 y^{2}

No Solution

Solution steps

Isolate

y

for

x = 3 y^{2} : y = \frac{x}{3}, y = - \frac{x}{3}

Isolate

y

for

x = 4 + 2 y^{2} : y = \frac{x - 4}{2}, y = - \frac{x - 4}{2}

f_{1} (x) = \frac{x}{3}

f_{2} (x) = - \frac{x}{3}

To find the intersection points solve

f_{i} (x) = f_{j} (x)

\frac{x}{3} = - \frac{x}{3} : x = 0

\frac{x}{3} = \frac{x - 4}{2} : x = 12

\frac{x}{3} = - \frac{x - 4}{2} :

No Solution for

x \in R

- \frac{x}{3} = \frac{x - 4}{2} :

No Solution for

x \in R

- \frac{x}{3} = - \frac{x - 4}{2} : x = 12

\frac{x - 4}{2} = - \frac{x - 4}{2} : x = 4

Curves not bounded

= No Solution

x \to - 4 lim (x^{2} - 3)

x^{2} y^{^{''}} + 3 x y^{^{'}} - 24 y = 0

\int \frac{4 x}{2 x + 3} d x

f (x) = 3 - 6 x

x \to 1 lim (- x + 2)

What is area x=3y^2,x=4+2y^2 ?
The answer to area x=3y^2,x=4+2y^2 is No Solution