What is volumeabout y=0,y=ln(x),y=0,[1,2] ?

The answer to volumeabout y=0,y=ln(x),y=0,[1,2] is 2piln^2(2)-4piln(2)+2pi

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volumeabout y=0,y=ln(x),y=0,[1,2]

volume about

y = 0, y = ln (x), y = 0, [1, 2]

2 π ln^{2} (2) - 4 π ln (2) + 2 π

Decimal

0.59161 \dots

Solution steps

Apply the volume formula:

\int_{1}^{2} π ∣ ln (x) ∣^{2} d x

Solve

\int_{1}^{2} π ∣ ln (x) ∣^{2} d x : 2 π ln^{2} (2) - 4 π ln (2) + 2 π

The volume for solid of revolution is:

= 2 π ln^{2} (2) - 4 π ln (2) + 2 π

\frac{d y}{d x} = - 9.81 + 0.00375 y^{2}

e^{- 3 t} cos (- t)

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\int \frac{x ^{2}}{( x ^{3} + 1 ) ^{2}} d x

What is volumeabout y=0,y=ln(x),y=0,[1,2] ?
The answer to volumeabout y=0,y=ln(x),y=0,[1,2] is 2piln^2(2)-4piln(2)+2pi