What is area x=3y^2,x=0,[0,2] ?

The answer to area x=3y^2,x=0,[0,2] is (8sqrt(2))/(3sqrt(3))

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area x=3y^2,x=0,[0,2]

area

x = 3 y^{2}, x = 0, [0, 2]

\frac{8 2}{3 3}

Decimal

2.17732 \dots

Solution steps

Isolate

y

for

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

Apply the area formula:

\int_{0}^{2} 2 \frac{x}{3} d x

Solve

\int_{0}^{2} 2 \frac{x}{3} d x : \frac{8 2}{3 3}

The area is:

= \frac{8 2}{3 3}

\int \frac{5 x ^{4} + 6}{( x ^{5} + 6 x ) ^{2}} d x

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\frac{d}{d x} (x^{4} sin (x))

\int 2 sin^{5} (x) cos^{- \frac{3}{2}} (x) d x

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

What is area x=3y^2,x=0,[0,2] ?
The answer to area x=3y^2,x=0,[0,2] is (8sqrt(2))/(3sqrt(3))