What is volumeabout y=0,x= 1/3 (y^2+2)^{3/2},[1,3] ?

The answer to volumeabout y=0,x= 1/3 (y^2+2)^{3/2},[1,3] is (pi(27*3^{1/3}-3*9^{1/3}-20))/5

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volumeabout y=0,x= 1/3 (y^2+2)^{3/2},[1,3]

volume about

y = 0, x = \frac{1}{3} (y^{2} + 2)^{\frac{3}{2}}, [1, 3]

\frac{π ( 27 \cdot 3 ^{\frac{1}{3}} - 3 \cdot 9 ^{\frac{1}{3}} - 20 )}{5}

Decimal Notation

7.97995 \dots

Solution steps

Apply the volume formula:

\int_{1}^{3} π (9^{\frac{1}{3}} x^{\frac{2}{3}} - 2) d x

Solve

\int_{1}^{3} π (9^{\frac{1}{3}} x^{\frac{2}{3}} - 2) d x : \frac{π ( 27 \cdot 3 ^{\frac{1}{3}} - 3 \cdot 9 ^{\frac{1}{3}} - 20 )}{5}

The volume for solid of revolution is:

= \frac{π ( 27 \cdot 3 ^{\frac{1}{3}} - 3 \cdot 9 ^{\frac{1}{3}} - 20 )}{5}

\frac{d y}{d x} = \frac{3 x}{10 y}

\int_{0}^{11} \frac{1}{121 + x ^{2}} d x

\int e^{2 x} x^{2} d x

\int \frac{2 x}{( 4 + 3 x ^{2} ) ^{2}} d x

3 cos (3 θ)

What is volumeabout y=0,x= 1/3 (y^2+2)^{3/2},[1,3] ?
The answer to volumeabout y=0,x= 1/3 (y^2+2)^{3/2},[1,3] is (pi(27*3^{1/3}-3*9^{1/3}-20))/5