What is area x=4y^3,x=22y^2 ?

The answer to area x=4y^3,x=22y^2 is 14641/48

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area x=4y^3,x=22y^2

area

x = 4 y^{3}, x = 22 y^{2}

\frac{14641}{48}

Decimal

305.02083 \dots

Solution steps

Isolate

y

for

x = 4 y^{3} : y = 3 \frac{x}{4}

Isolate

y

for

x = 22 y^{2} : y = \frac{x}{22}, y = - \frac{x}{22}

f_{1} (x) = 3 \frac{x}{4}

f_{2} (x) = \frac{x}{22}

To find the intersection points solve

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

3 \frac{x}{4} = \frac{x}{22} : x = 0, x = \frac{1331}{2}

3 \frac{x}{4} = - \frac{x}{22} : x = 0

\frac{x}{22} = - \frac{x}{22} : x = 0

= \int_{0}^{\frac{1331}{2}} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{0}^{\frac{1331}{2}} 3 \frac{x}{4} - \frac{x}{22} d x

\int_{0}^{\frac{1331}{2}} 3 \frac{x}{4} - \frac{x}{22} d x = \frac{14641}{48}

= \frac{14641}{48}

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