Math Problem Statement

Find the volume of the solid obtained by rotating the region bounded by the curves y = 4√x - 2, y = 0, y = 5 about the y-axis.

Solution

We are tasked with finding the volume of a solid obtained by rotating a region about the yy-axis. The region is bounded by the curves:

y=4x2,y=0,y=5y = 4\sqrt{x} - 2, \quad y = 0, \quad y = 5

Part (a) - Definite Integral Setup

Since the rotation is about the yy-axis, we should express xx as a function of yy. From the equation y=4x2y = 4\sqrt{x} - 2, solve for xx:

y+2=4xy + 2 = 4\sqrt{x} x=y+24\sqrt{x} = \frac{y + 2}{4} x=(y+24)2=(y+2)216x = \left( \frac{y + 2}{4} \right)^2 = \frac{(y + 2)^2}{16}

The formula for the volume of a solid of revolution about the yy-axis is given by the method of cylindrical shells:

V=2πy1y2xdyV = 2\pi \int_{y_1}^{y_2} x \, dy

In this case, x=(y+2)216x = \frac{(y + 2)^2}{16}, and the limits of integration for yy are from 00 to 55. Therefore, the definite integral to find the volume is:

V=2π05(y+2)216dyV = 2\pi \int_0^5 \frac{(y + 2)^2}{16} \, dy

Part (b) - Compute the Volume

Now, let's compute the integral:

V=2π11605(y+2)2dyV = 2\pi \cdot \frac{1}{16} \int_0^5 (y + 2)^2 \, dy

First, expand (y+2)2(y + 2)^2:

(y+2)2=y2+4y+4(y + 2)^2 = y^2 + 4y + 4

Now the integral becomes:

V=π805(y2+4y+4)dyV = \frac{\pi}{8} \int_0^5 (y^2 + 4y + 4) \, dy

Now we can compute this integral step by step:

05y2dy=[y33]05=533=1253\int_0^5 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^5 = \frac{5^3}{3} = \frac{125}{3} 054ydy=[2y2]05=2×25=50\int_0^5 4y \, dy = \left[ 2y^2 \right]_0^5 = 2 \times 25 = 50 054dy=[4y]05=4×5=20\int_0^5 4 \, dy = \left[ 4y \right]_0^5 = 4 \times 5 = 20

Adding all these up:

V=π8(1253+50+20)V = \frac{\pi}{8} \left( \frac{125}{3} + 50 + 20 \right) V=π8×(125+150+603)=π8×3353V = \frac{\pi}{8} \times \left( \frac{125 + 150 + 60}{3} \right) = \frac{\pi}{8} \times \frac{335}{3} V=335π24V = \frac{335\pi}{24}

Now, I'll compute the numerical value of this expression.The volume of the solid obtained by rotating the given region about the yy-axis is approximately 43.85139743.851397 cubic units, rounded to six decimal places.

Do you have any questions or would you like further details on any part of the solution?

Here are five related questions for further exploration:

  1. How would the integral change if the region was rotated around the x-axis instead of the y-axis?
  2. What happens if the upper limit of yy is changed from 5 to another value?
  3. How does the method of cylindrical shells differ from the method of disks or washers?
  4. Can you generalize this approach for a function y=axby = a\sqrt{x} - b?
  5. What if the curve y=4x2y = 4\sqrt{x} - 2 was bounded by different lines?

Tip: When solving problems involving volumes of revolution, always ensure the function is correctly expressed in terms of the variable corresponding to the axis of rotation.

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Math Problem Analysis

Mathematical Concepts

Solid of revolution
Cylindrical shells
Integration
Algebraic manipulation

Formulas

V = 2π ∫(x dy) from y1 to y2
x = ((y + 2) / 4)^2

Theorems

Volume of revolution using the cylindrical shell method

Suitable Grade Level

Undergraduate (Calculus II level)