y = -0.0159x* + 0.2301x3 1.2034x2 + 2.6595x + 0.0661 What is the volume of this solid of revolution bounded between x=0 and x=6 ?

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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How do I solve this using integration?

**Problem Statement**

Given the function:

\[ y = -0.0159x^4 + 0.2301x^3 - 1.2034x^2 + 2.6595x + 0.0661 \]

Determine the volume of the solid of revolution formed when this curve is revolved around the x-axis, bounded between \( x = 0 \) and \( x = 6 \).

**Explanation**

To find the volume of the solid of revolution, use the disk method (or the method of disks). The formula for the volume \( V \) of a solid of revolution about the x-axis is:

\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]

In this problem, the limits of integration are \( a = 0 \) and \( b = 6 \), and the function \( f(x) \) is the given polynomial function of \( x \).
Transcribed Image Text:**Problem Statement** Given the function: \[ y = -0.0159x^4 + 0.2301x^3 - 1.2034x^2 + 2.6595x + 0.0661 \] Determine the volume of the solid of revolution formed when this curve is revolved around the x-axis, bounded between \( x = 0 \) and \( x = 6 \). **Explanation** To find the volume of the solid of revolution, use the disk method (or the method of disks). The formula for the volume \( V \) of a solid of revolution about the x-axis is: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] In this problem, the limits of integration are \( a = 0 \) and \( b = 6 \), and the function \( f(x) \) is the given polynomial function of \( x \).
Expert Solution
Step 1

We know that Volume of solid formed by revolving the region bounded by the curve y=f(x) and the x-axis between x= a & x=b about the x-axis is V=πab[f(x)]2dx

Here  f(x)=-0.0159x4+0.2301x3-1.2034x2+2.6595x+0.0661    & revolution is b/w x= 0 & x=6

so, Volume of solid of revolution about x-axis is =π06[-0.0159x4+0.2301x3-1.2034x2+2.6595x+0.0661]2dx

                                                                            =π06[0.00025x8+0.05294x6+1.44817x4+7.07214x2+0.00436+2(-0.0159)(0.2301)x7+2(-0.0159)(-1.2034)x6                      +2(-0.0159)(2.6595)x5+2(-0.0159)(0.0661)x4+2(0.2301)(-1.2034)x5+2(0.2301)(2.6595)x4+2(0.2301)(0.0661)x3+                      + 2(-1.2034)(2.6595)x3+2(-1.2034)(0.0661)x2+2(2.6595)(0.0661)x]dx=π06[0.00025x8+0.05294x6+1.44817x4+7.07214x2+0.00436-0.00731x7+0.03826x6-0.08457x5-0.00210x4-0.55380x5                         +1.22390x4+0.03041x3-6.40088x3-0.15908x2+0.35158x]dx=π06[0.00025x8-0.00731x7+0.0912x6-0.63837x5+2.67207x4-6.37047x3+6.91306x2+0.35158x+0.00436]dx=π0.00025x99-0.00731x88+0.0912x77-0.63837x66+2.67207x55-6.37047x44+6.91306x33+0.35158x22+0.00436x06=π0.00025(6)99-0.00731(6)88+0.0912(6)77-0.63837(6)66+2.67207(6)55-6.37047(6)44+6.91306(6)33+0.35158(6)22+0.00436(6)=3.14[279.936-1534.749+3647.166-4963.965+4155.603-2064.032+497.740+6.328+0.0261]=3.14[24.056]=75.536 cubic units

                                                                           

  

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