What is area x=2y^2,x=4-y^2 ?

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

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area x=2y^2,x=4-y^2

area

x = 2 y^{2}, x = 4 - y^{2}

No Solution

Solution steps

Isolate

y

for

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

Isolate

y

for

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

f_{1} (x) = \frac{x}{2}

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

To find the intersection points solve

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

\frac{x}{2} = - \frac{x}{2} : x = 0

\frac{x}{2} = - x + 4 : x = \frac{8}{3}

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

No Solution for

x \in R

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

No Solution for

x \in R

- \frac{x}{2} = - - x + 4 : x = \frac{8}{3}

- x + 4 = - - x + 4 : x = 4

Curves not bounded

= No Solution

\frac{\partial}{\partial x} (\frac{z}{x})

\int \frac{( x ^{3} + 10 )}{x ^{3} - 2 x ^{2} + 5 x} d x

\int - cos (y) d y

n = 0 \sum \infty 4 \cdot (- \frac{1}{8})^{n}

y \to - 2 lim (\frac{y ^{3} + 8}{y + 2})