Find the volume of the spindle shaped solid generated by revolving astroid

x

2

3 + y

2

3 = a

2

3 about the x-axis.

To find the volume of the spindle-shaped solid generated by revolving the astroid \( x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}} \) about the x-axis, we can use the method of disks or washers. Here's the detailed process:

1. **Equation of the Astroid:**
   The given equation is:
   \[
   x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}
   \]
   
2. **Solving for \( y \) in terms of \( x \):**
   Isolate \( y \):
   \[
   y = \left(a^{\frac{2}{3}} - x^{\frac{2}{3}}\right)^{\frac{3}{2}}
   \]

3. **Volume of Revolution Using the Disk Method:**
   When revolving around the x-axis, the volume \( V \) can be found using the formula:
   \[
   V = \pi \int_{-a}^{a} \left[ y(x) \right]^2 dx
   \]
   Here, \( y(x) = \left(a^{\frac{2}{3}} - x^{\frac{2}{3}}\right)^{\frac{3}{2}} \).

4. **Setting up the Integral:**
   \[
   V = \pi \int_{-a}^{a} \left( \left(a^{\frac{2}{3}} - x^{\frac{2}{3}}\right)^{\frac{3}{2}} \right)^2 dx
   \]
   Simplify the integrand:
   \[
   V = \pi \int_{-a}^{a} \left(a^{\frac{2}{3}} - x^{\frac{2}{3}}\right)^3 dx
   \]

5. **Symmetry Consideration:**
   Because the function is symmetric about the y-axis, we can compute the integral from 0 to \( a \) and then double the result:
   \[
   V = 2\pi \int_{0}^{a} \left(a^{\frac{2}{3}} - x^{\frac{2}{3}}\right)^3 dx
   \]

6. **Substitute and Integrate:**
   Let \( x = a \cos^3 \theta \). Then \( dx = -3a \cos^2 \theta \sin \theta d\theta \).
   When \( x = 0 \), \( \theta = \frac{\pi}{2} \).
   When \( x = a \), \( \theta = 0 \).

   The integral becomes:
   \[
   V = 2\pi \int_{0}^{\frac{\pi}{2}} \left(a^{\frac{2}{3}} - (a \cos^3 \theta)^{\frac{2}{3}}\right)^3 \cdot (-3a \cos^2 \theta \sin \theta) d\theta
   \]

   Simplifying:
   \[
   V = 6\pi a \int_{0}^{\frac{\pi}{2}} (1 - \cos^2 \theta)^3 \cos^2 \theta \sin \theta d\theta
   \]

7. **Simplify the Integrand:**
   Use the identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
   \[
   (1 - \cos^2 \theta)^3 = \sin^6 \theta
   \]

   The integral now is:
   \[
   V = 6\pi a \int_{0}^{\frac{\pi}{2}} \sin^6 \theta \cos^2 \theta \sin \theta d\theta
   \]

8. **Combine Terms and Integrate:**
   \[
   V = 6\pi a \int_{0}^{\frac{\pi}{2}} \sin^7 \theta \cos^2 \theta d\theta
   \]

   This can be solved using a beta function or standard integral tables.

   \[
   \int_{0}^{\frac{\pi}{2}} \sin^7 \theta \cos^2 \theta d\theta = \frac{8}{315}
   \]

   So,
   \[
   V = 6\pi a \cdot \frac{8}{315} = \frac{48\pi a}{315} = \frac{16\pi a}{105}
   \]

9. **Final Volume:**
   \[
   V = \frac{16\pi a^3}{105}
   \]

Thus, the volume of the spindle-shaped solid is:
\[
V = \frac{16\pi a^3}{105}
\]

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. How to find the volume of a solid of revolution using the shell method?
2. What are the applications of astroids in physics or engineering?
3. How to compute the volume of a torus?
4. What is the difference between the disk method and the washer method in calculus?
5. How to find the surface area of a solid of revolution?

**Tip:** When integrating functions involving powers of sine and cosine, using trigonometric identities can often simplify the process.

Learn how to calculate the volume of a spindle-shaped solid formed by revolving the astroid x^(2/3) + y^(2/3) = a^(2/3) about the x-axis. Detailed step-by-step solution using the disk method in calculus.

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