Math Problem Statement

solution step by step

Evaluate the triple integral \( \iiint_V (x^2 + y^2 + z^2) \, dV \) over the volume bounded by the planes \( z = 0 \), \( z = 1 \), and the cylinder \( x^2 + y^2 = 1 \).

Solution

To solve the triple integral:

V(x2+y2+z2)dV\iiint_V (x^2 + y^2 + z^2) \, dV

over the volume bounded by the planes z=0z = 0, z=1z = 1, and the cylinder x2+y2=1x^2 + y^2 = 1, we can follow these steps:

Step 1: Set up the coordinates

Given the cylindrical symmetry of the region (bounded by x2+y2=1x^2 + y^2 = 1), it is convenient to switch to cylindrical coordinates where:

  • x=rcosθx = r \cos \theta
  • y=rsinθy = r \sin \theta
  • z=zz = z
  • dV=rdrdθdzdV = r \, dr \, d\theta \, dz

In cylindrical coordinates, the integrand x2+y2+z2x^2 + y^2 + z^2 becomes: x2+y2+z2=r2+z2x^2 + y^2 + z^2 = r^2 + z^2

Step 2: Set up the bounds

The volume is bounded by:

  1. z=0z = 0 (bottom plane),
  2. z=1z = 1 (top plane), and
  3. x2+y2=1x^2 + y^2 = 1 (cylinder), which implies 0r10 \leq r \leq 1.

Thus, the bounds for each variable in cylindrical coordinates are:

  • 0r10 \leq r \leq 1
  • 0θ2π0 \leq \theta \leq 2\pi
  • 0z10 \leq z \leq 1

Step 3: Rewrite the integral in cylindrical coordinates

Substitute the integrand and bounds into the integral:

V(x2+y2+z2)dV=02π0101(r2+z2)rdzdrdθ\iiint_V (x^2 + y^2 + z^2) \, dV = \int_0^{2\pi} \int_0^1 \int_0^1 (r^2 + z^2) \cdot r \, dz \, dr \, d\theta

Step 4: Separate the integral

Now, distribute rr and separate the integral:

=02π0101(r3+rz2)dzdrdθ= \int_0^{2\pi} \int_0^1 \int_0^1 (r^3 + r z^2) \, dz \, dr \, d\theta

Step 5: Integrate with respect to zz

Now, integrate the inner integral with respect to zz:

=02π01[r3z+rz33]01drdθ= \int_0^{2\pi} \int_0^1 \left[ r^3 z + r \frac{z^3}{3} \right]_0^1 \, dr \, d\theta =02π01(r3+r3)drdθ= \int_0^{2\pi} \int_0^1 \left( r^3 + \frac{r}{3} \right) \, dr \, d\theta

Step 6: Integrate with respect to rr

Next, integrate with respect to rr:

=02π[r44+r26]01dθ= \int_0^{2\pi} \left[ \frac{r^4}{4} + \frac{r^2}{6} \right]_0^1 \, d\theta =02π(14+16)dθ= \int_0^{2\pi} \left( \frac{1}{4} + \frac{1}{6} \right) \, d\theta =02π512dθ= \int_0^{2\pi} \frac{5}{12} \, d\theta

Step 7: Integrate with respect to θ\theta

Finally, integrate with respect to θ\theta:

=5122π=5π6= \frac{5}{12} \cdot 2\pi = \frac{5\pi}{6}

Final Answer

The value of the integral is:

5π6\frac{5\pi}{6}

Would you like any further explanation on any of these steps?


Here are some related questions that could deepen your understanding:

  1. How would the solution change if the upper bound for zz was z=2z = 2 instead of z=1z = 1?
  2. What would be the result if the region was bounded by a sphere instead of a cylinder?
  3. Can we solve this problem using spherical coordinates instead, and would it simplify the integral?
  4. How do cylindrical coordinates help simplify problems with cylindrical symmetry?
  5. What other types of regions would be suited for cylindrical coordinates?

Tip: Converting integrals to cylindrical or spherical coordinates can significantly simplify integrals involving regions with circular or spherical symmetry.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Triple Integrals
Cylindrical Coordinates
Volume Integrals

Formulas

Cylindrical coordinates: \( x = r \cos \theta \), \( y = r \sin \theta \), \( z = z \)
Volume element in cylindrical coordinates: \( dV = r \, dr \, d\theta \, dz \)

Theorems

Triple Integral Theorem

Suitable Grade Level

Undergraduate Calculus