CALCU-Calculate the volume of the solid obtained b

Calculate the volume of the solid obtained by rotating the region enclosed by

y = x^3

,

y = 0

,

x = 0

, and

x = 2

around the x-axis.

Generated Graph

Solution by Steps

step 1

To find the volume of the solid obtained by rotating the region enclosed by

y = x^3

,

y = 0

,

x = 0

, and

x = 2

around the x-axis, we use the disk method. The volume

V

is given by the integral:

V = \pi \int_0^2 (x^3)^2 \, dx

step 2

Simplifying the integral, we have:

V = \pi \int_0^2 x^6 \, dx

step 3

Evaluating the integral:

V = \pi \left[ \frac{x^7}{7} \right]_0^2 = \pi \left( \frac{2^7}{7} - 0 \right) = \pi \left( \frac{128}{7} \right)

step 4

Therefore, the volume of the solid is:

V = \frac{128\pi}{7} \approx 57.446

Answer

The volume of the solid is

\frac{128\pi}{7}

or approximately 57.446 cubic units.

Key Concept

The disk method is used to find the volume of solids of revolution by integrating the area of circular disks.

Explanation

The volume is calculated by integrating the area of disks formed by rotating the function around the x-axis, leading to the final result.

Solution by Steps

step 1

To find the derivative of the expression given, we start with the expression:

\\frac{d}{d}=\\left(\\int \\ln \\left(\\int\\right)\\right) 2 z \\ln (z)+\\int\\left(\\frac{1}{z}\\right)

step 2

We need to differentiate each part of the expression. The first part involves the integral of the logarithm, and the second part is the integral of

\\frac{1}{z}

step 3

The derivative of the first part can be computed using the Fundamental Theorem of Calculus, while the second part simplifies to

\\ln(z)

step 4

Therefore, the overall derivative is given by combining the results from the differentiation of both parts

Answer

The derivative of the expression is

\\frac{d}{d} = 2z \\ln(z) + \\ln(z)

Key Concept

Understanding how to differentiate integrals and logarithmic functions is crucial in calculus.

Explanation

The answer combines the results of differentiating the integral expressions, applying the Fundamental Theorem of Calculus and properties of logarithms.

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