What is volumeabout x=0,y=(3x+1)^{1/4},[0,8] ?

The answer to volumeabout x=0,y=(3x+1)^{1/4},[0,8] is (670499176pi)/(405)

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volumeabout x=0,y=(3x+1)^{1/4},[0,8]

volume about

x = 0, y = (3 x + 1)^{\frac{1}{4}}, [0, 8]

\frac{670499176 π}{405}

Decimal Notation

5201074.77915 \dots

Solution steps

Apply the volume formula:

\int_{0}^{8} \frac{π}{9} (y^{4} - 1)^{2} d y

Solve

\int_{0}^{8} \frac{π}{9} (y^{4} - 1)^{2} d y : \frac{670499176 π}{405}

The volume for solid of revolution is:

= \frac{670499176 π}{405}

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

2 x^{2} - 7 x + 5

(x + \frac{4}{x})^{^{'}}

\frac{\partial}{\partial x} (e^{- (x^{2} + y^{2} + z^{2})})

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

What is volumeabout x=0,y=(3x+1)^{1/4},[0,8] ?
The answer to volumeabout x=0,y=(3x+1)^{1/4},[0,8] is (670499176pi)/(405)