Math Problem Statement

how is it 7.2649330114264

Solution

Let's go through the integration steps again to verify the calculations.

The integral setup for the volume is: V=18π78π22.55z3rdrdzdθV = \int_{\frac{1}{8}\pi}^{\frac{7}{8}\pi} \int_{2}^{2.5} \int_{5-z}^{3} r \, dr \, dz \, d\theta

  1. Integrate with respect to rr: 5z3rdr=[r22]5z3=92(5z)22=922510z+z22=925+10zz22=16+10zz22=8+5zz22\int_{5-z}^{3} r \, dr = \left[ \frac{r^2}{2} \right]_{5-z}^{3} = \frac{9}{2} - \frac{(5-z)^2}{2} = \frac{9}{2} - \frac{25 - 10z + z^2}{2} = \frac{9 - 25 + 10z - z^2}{2} = \frac{-16 + 10z - z^2}{2} = -8 + 5z - \frac{z^2}{2}

  2. Integrate with respect to zz: 22.5(8+5zz22)dz\int_{2}^{2.5} (-8 + 5z - \frac{z^2}{2}) \, dz =[8z+5z22z36]22.5= \left[ -8z + \frac{5z^2}{2} - \frac{z^3}{6} \right]_{2}^{2.5} Evaluating the antiderivative at the bounds:

For z=2.5z = 2.5: 8(2.5)+5(2.5)22(2.5)36=20+56.25215.6256-8(2.5) + \frac{5(2.5)^2}{2} - \frac{(2.5)^3}{6} = -20 + \frac{5 \cdot 6.25}{2} - \frac{15.625}{6} =20+31.2522.6042= -20 + \frac{31.25}{2} - 2.6042 =20+15.6252.6042=6.9792= -20 + 15.625 - 2.6042 = -6.9792

For z=2z = 2: 8(2)+5(2)22236=16+54286-8(2) + \frac{5(2)^2}{2} - \frac{2^3}{6} = -16 + \frac{5 \cdot 4}{2} - \frac{8}{6} =16+101.3333=7.3333= -16 + 10 - 1.3333 = -7.3333

So, the definite integral is: 6.9792(7.3333)=6.9792+7.3333=0.3541-6.9792 - (-7.3333) = -6.9792 + 7.3333 = 0.3541

  1. Integrate with respect to θ\theta: 18π78π0.3541dθ=0.3541(78π18π)\int_{\frac{1}{8}\pi}^{\frac{7}{8}\pi} 0.3541 \, d\theta = 0.3541 \left( \frac{7}{8}\pi - \frac{1}{8}\pi \right) =0.354168π=0.35410.75π= 0.3541 \cdot \frac{6}{8} \pi = 0.3541 \cdot 0.75 \pi =0.265575π= 0.265575 \pi

Evaluating the final numerical value: 0.265575π0.2655753.141590.83430.265575 \pi \approx 0.265575 \cdot 3.14159 \approx 0.8343

Therefore, the volume of the solid is approximately 0.8343cubic units0.8343 \, \text{cubic units}.

The previous result of 7.2649330114264 seems to be incorrect based on this re-evaluation.

Do you have any more questions or need further details?

Here are 5 numbered questions you can ask:

  1. Can you explain how to integrate with respect to rr in cylindrical coordinates?
  2. How do cylindrical coordinates simplify solving volume integrals?
  3. What is the physical significance of the volume element in cylindrical coordinates?
  4. Can you provide examples of similar volume integration problems?
  5. How can we verify the results of such integrals?

Tip: Always double-check each step of integration, especially when converting bounds and evaluating definite integrals, to ensure accuracy.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integral calculus
Volume calculation
Cylindrical coordinates

Formulas

Volume integral formula in cylindrical coordinates

Theorems

-

Suitable Grade Level

Advanced College