Math Problem Statement

find the volume of the solid y = 2cos(x^2) , y = 0 , x= 0 , x = sqrt pi/4, rotated about the y axis

Solution

To find the volume of the solid formed by rotating the region between the curve y=2cos(x2)y = 2\cos(x^2), y=0y = 0, and the vertical lines x=0x = 0 and x=π/4x = \sqrt{\pi}/4 about the y-axis, we can use the method of cylindrical shells.

The formula for the volume of a solid obtained by rotating a region about the y-axis is given by:

V=2πabxf(x)dxV = 2\pi \int_{a}^{b} x \cdot f(x) \, dx

Here:

  • f(x)=2cos(x2)f(x) = 2\cos(x^2) is the function that defines the curve.
  • a=0a = 0 and b=π/4b = \sqrt{\pi}/4 are the bounds for xx.
  • xx is the radius of the cylindrical shell at a given point.
  • The integral gives the volume of the cylindrical shells between these bounds.

Step 1: Set up the integral

The volume is:

V=2π0π/4x2cos(x2)dxV = 2\pi \int_{0}^{\sqrt{\pi}/4} x \cdot 2\cos(x^2) \, dx

This simplifies to:

V=4π0π/4xcos(x2)dxV = 4\pi \int_{0}^{\sqrt{\pi}/4} x \cos(x^2) \, dx

Step 2: Substitution

We can solve the integral using the substitution u=x2u = x^2, so that du=2xdxdu = 2x \, dx, which means xdx=du2x \, dx = \frac{du}{2}.

  • When x=0x = 0, u=0u = 0.
  • When x=π/4x = \sqrt{\pi}/4, u=π/16u = \pi/16.

Thus, the integral becomes:

V=4π0π/16cos(u)du2V = 4\pi \int_{0}^{\pi/16} \cos(u) \cdot \frac{du}{2}

Simplifying:

V=2π0π/16cos(u)duV = 2\pi \int_{0}^{\pi/16} \cos(u) \, du

Step 3: Integrating

The integral of cos(u)\cos(u) is sin(u)\sin(u), so we have:

V=2π[sin(u)]0π/16V = 2\pi [\sin(u)]_{0}^{\pi/16}

Evaluating:

V=2π(sin(π16)sin(0))V = 2\pi \left( \sin\left(\frac{\pi}{16}\right) - \sin(0) \right)

V=2πsin(π16)V = 2\pi \sin\left(\frac{\pi}{16}\right)

Step 4: Final Answer

Thus, the volume is:

V=2πsin(π16)V = 2\pi \sin\left(\frac{\pi}{16}\right)

If you need a numerical approximation:

sin(π16)0.195\sin\left(\frac{\pi}{16}\right) \approx 0.195

So:

V2π×0.1951.225V \approx 2\pi \times 0.195 \approx 1.225

Summary

The volume of the solid is approximately 1.225cubic units1.225 \, \text{cubic units}.

Would you like more details or have any further questions? Here are some related questions:

  1. How does the method of cylindrical shells differ from the disk method in volume calculation?
  2. What is the significance of using substitution in integration?
  3. How would the volume change if the region was rotated about the x-axis instead of the y-axis?
  4. Can we compute volumes for regions with parametric equations using similar methods?
  5. How would you approach a volume problem involving a more complex function, like an exponential or logarithmic one?

Tip: When dealing with integrals involving trigonometric functions and substitutions, check for standard identities or symmetry that could simplify the process!

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Math Problem Analysis

Mathematical Concepts

Calculus
Volume of Solids
Cylindrical Shells
Integration

Formulas

V = 2π ∫(a to b) x * f(x) dx
Substitution method for integration

Theorems

Method of cylindrical shells

Suitable Grade Level

Grades 11-12 (Advanced Calculus)