en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area x=4y^2,x=8+2y^2

Solution

area

x = 4 y^{2}, x = 8 + 2 y^{2}

Solution

No Solution

Solution steps

Isolate

y

for

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

Isolate

y

for

x = 8 + 2 y^{2} : y = \frac{x - 8}{2}, y = - \frac{x - 8}{2}

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} = \frac{x - 8}{2} : x = 16

\frac{x}{2} = - \frac{x - 8}{2} :

No Solution for

x \in R

- \frac{x}{2} = \frac{x - 8}{2} :

No Solution for

x \in R

- \frac{x}{2} = - \frac{x - 8}{2} : x = 16

\frac{x - 8}{2} = - \frac{x - 8}{2} : x = 8

Curves not bounded

= No Solution

Popular Examples

derivative of 3x^2e^{x^3}

\frac{d}{d x} (3 x^{2} e^{x^{3}})

t^{2} y^{^{'}} + 2 t y = 3

integral of 3/(x(x^2+25))

\int \frac{3}{x ( x ^{2} + 25 )} d x

simplify 6x^3+7/x simplify

6 x^{3} + \frac{7}{x}

(4x^2)y^{''}+4xy^'=0

(4 x^{2}) y^{^{''}} + 4 x y^{^{'}} = 0

Frequently Asked Questions (FAQ)

What is area x=4y^2,x=8+2y^2 ?
The answer to area x=4y^2,x=8+2y^2 is No Solution