Math Problem Statement

The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method. y = −x2 + 10x − 16, y = 0; about the x-axis

Solution

We are tasked with finding the volume of the solid generated by rotating the region bounded by the curves y=x2+10x16y = -x^2 + 10x - 16 and y=0y = 0 (the x-axis) around the x-axis. This can be solved using the disk method, as we are revolving around the x-axis.

Step 1: Find the intersection points

We first need to determine where the parabola intersects the x-axis, i.e., where y=0y = 0.

The given equation is:

y=x2+10x16=0y = -x^2 + 10x - 16 = 0

Solve for xx by applying the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1a = -1, b=10b = 10, and c=16c = -16.

x=10±1024(1)(16)2(1)x = \frac{-10 \pm \sqrt{10^2 - 4(-1)(-16)}}{2(-1)} x=10±100642x = \frac{-10 \pm \sqrt{100 - 64}}{-2} x=10±362x = \frac{-10 \pm \sqrt{36}}{-2} x=10±62x = \frac{-10 \pm 6}{-2} This gives us two solutions:

x=10+62=2andx=1062=8x = \frac{-10 + 6}{-2} = 2 \quad \text{and} \quad x = \frac{-10 - 6}{-2} = 8

So, the region is bounded between x=2x = 2 and x=8x = 8.

Step 2: Set up the integral using the disk method

For the disk method, the volume VV is given by:

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

In this case, f(x)=x2+10x16f(x) = -x^2 + 10x - 16, and we are integrating from x=2x = 2 to x=8x = 8. Thus, the volume is:

V=π28[x2+10x16]2dxV = \pi \int_{2}^{8} [-x^2 + 10x - 16]^2 \, dx

Step 3: Compute the integral

Now, let's compute the integral:

V=π28(x2+10x16)2dxV = \pi \int_{2}^{8} (-x^2 + 10x - 16)^2 \, dx

We can expand the integrand:

(x2+10x16)2=x420x3+132x2320x+256(-x^2 + 10x - 16)^2 = x^4 - 20x^3 + 132x^2 - 320x + 256

Now the integral becomes:

V=π28(x420x3+132x2320x+256)dxV = \pi \int_{2}^{8} (x^4 - 20x^3 + 132x^2 - 320x + 256) \, dx

Integrating term by term:

V=π[x555x4+44x3160x2+256x]28V = \pi \left[ \frac{x^5}{5} - 5x^4 + 44x^3 - 160x^2 + 256x \right]_{2}^{8}

Step 4: Evaluate the definite integral

Now, substitute the limits of integration x=8x = 8 and x=2x = 2:

  1. For x=8x = 8: 8555(84)+44(83)160(82)+256(8)=3276855(4096)+44(512)160(64)+256(8)\frac{8^5}{5} - 5(8^4) + 44(8^3) - 160(8^2) + 256(8) = \frac{32768}{5} - 5(4096) + 44(512) - 160(64) + 256(8) =6553.620480+2252810240+2048=6409.6= 6553.6 - 20480 + 22528 - 10240 + 2048 = 6409.6

  2. For x=2x = 2: 2555(24)+44(23)160(22)+256(2)=3255(16)+44(8)160(4)+256(2)\frac{2^5}{5} - 5(2^4) + 44(2^3) - 160(2^2) + 256(2) = \frac{32}{5} - 5(16) + 44(8) - 160(4) + 256(2) =6.480+352640+512=150.4= 6.4 - 80 + 352 - 640 + 512 = 150.4

Now subtract the two values:

6409.6150.4=6259.26409.6 - 150.4 = 6259.2

Step 5: Multiply by π\pi

Finally, multiply the result by π\pi:

V=6259.2πcubic unitsV = 6259.2\pi \, \text{cubic units}

Thus, the volume of the solid is approximately:

V6259.2πV \approx 6259.2\pi

Do you want more details or have any questions?

Here are five related questions to explore further:

  1. How would the volume change if the region were rotated about the y-axis instead of the x-axis?
  2. Can the volume be found using the shell method? If so, how?
  3. What if the equation of the curve were shifted vertically? How would that affect the volume?
  4. How does the disk method differ from the washer method, and when should each be used?
  5. Can the same solid volume be computed using a different coordinate system?

Tip: Always check the limits of integration when solving for volumes to ensure you're considering the correct bounds.

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Math Problem Analysis

Mathematical Concepts

Calculus
Volume of Solids of Revolution
Disk Method
Quadratic Equations

Formulas

Volume using the disk method: V = π ∫[a, b] [f(x)]^2 dx
Quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Quadratic formula
Disk method for solids of revolution

Suitable Grade Level

Grades 11-12 and above (Advanced Calculus)