Math Problem Statement

find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis

be aware: rotation about the "x-axis"

shell method with respect to dx

answer should be 512pi/5

full process, thanks

Find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis using the shell method with respect to dx.

Solution

To find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis using the shell method, follow these steps carefully.


Step 1: Understanding the Given Equation

We are given:

x=4y2y3x = 4y^2 - y^3

and the boundary condition:

x=0x = 0

This means we need to solve for values of yy where:

4y2y3=04y^2 - y^3 = 0

Step 2: Solving for yy

Factor the equation:

y2(4y)=0y^2(4 - y) = 0

This gives:

y2=0or4y=0y^2 = 0 \quad \text{or} \quad 4 - y = 0

y=0ory=4y = 0 \quad \text{or} \quad y = 4

Thus, the region is bounded by y=0y = 0 and y=4y = 4.


Step 3: Shell Method Formula

The shell method formula for rotation about the x-axis is:

V=2πab(radius)(height)dyV = 2\pi \int_{a}^{b} (\text{radius}) (\text{height}) \, dy

  • Radius = yy (distance from the shell to the x-axis)
  • Height = x=4y2y3x = 4y^2 - y^3 (horizontal width of the shell)
  • Limits of Integration: y=0y = 0 to y=4y = 4

Thus, the volume integral is:

V=2π04y(4y2y3)dyV = 2\pi \int_0^4 y (4y^2 - y^3) \, dy


Step 4: Computing the Integral

Expanding the terms inside the integral:

V=2π04(4y3y4)dyV = 2\pi \int_0^4 (4y^3 - y^4) \, dy

Now, integrate term by term:

044y3dy=4y4404=y404=4404=2560=256\int_0^4 4y^3 \, dy = 4 \cdot \frac{y^4}{4} \Big|_0^4 = y^4 \Big|_0^4 = 4^4 - 0^4 = 256 - 0 = 256

04y4dy=y5504=455055=102450=10245\int_0^4 y^4 \, dy = \frac{y^5}{5} \Big|_0^4 = \frac{4^5}{5} - \frac{0^5}{5} = \frac{1024}{5} - 0 = \frac{1024}{5}


Step 5: Evaluating the Expression

V=2π(25610245)V = 2\pi \left( 256 - \frac{1024}{5} \right)

Find a common denominator:

256=12805256 = \frac{1280}{5}

V=2π(1280510245)V = 2\pi \left( \frac{1280}{5} - \frac{1024}{5} \right)

V=2π×2565V = 2\pi \times \frac{256}{5}

V=512π5V = \frac{512\pi}{5}


Final Answer:

512π5\boxed{\frac{512\pi}{5}}

Do you have any questions, or would you like more details?

Related Questions:

  1. How does the shell method differ from the disk/washer method?
  2. What happens if the region is rotated around the y-axis instead?
  3. How do you determine when to use shells versus washers?
  4. What if the function were x=4y2+y3x = 4y^2 + y^3? How would the volume change?
  5. How would this problem change if rotated around y=5y = 5?

Tip:

For the shell method, always set up the radius and height carefully, as they depend on the axis of rotation.

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Math Problem Analysis

Mathematical Concepts

Volume of solids of revolution
Shell method
Definite integrals

Formulas

V = 2π ∫ (radius)(height) dy
Integration of polynomial functions: ∫ x^n dx = x^(n+1) / (n+1)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12, Calculus