what is ythe volume of the solid bounded by z=-2y, the xy plane, r=5costheta

To find the volume of the solid bounded by the surfaces \( z = -2y \), the \( xy \)-plane, and the cylinder \( r = 5\cos\theta \), we use cylindrical coordinates where \( r \), \( \theta \), and \( z \) represent the radial distance, angle, and height, respectively.

### Step 1: Express the surfaces in cylindrical coordinates
- The surface \( z = -2y \) in cylindrical coordinates becomes \( z = -2r\sin\theta \) because \( y = r\sin\theta \).
- The \( xy \)-plane corresponds to \( z = 0 \).
- The curve \( r = 5\cos\theta \) is a cardioid in the polar plane.

### Step 2: Determine the bounds for \( r \) and \( \theta \)
- Since \( r = 5\cos\theta \) bounds the region, \( r \) ranges from 0 to \( 5\cos\theta \).
- \( \theta \) ranges from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) because \( 5\cos\theta \) is positive for this interval.

### Step 3: Set up the volume integral
The volume is given by the triple integral:
\[
V = \int_{\theta=-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{r=0}^{5\cos\theta} \int_{z=0}^{-2r\sin\theta} r \, dz \, dr \, d\theta
\]
Here, the extra \( r \) in the integrand accounts for the Jacobian when converting to cylindrical coordinates.

### Step 4: Evaluate the integral
1. **Integrate with respect to \( z \):**
   \[
   \int_{z=0}^{-2r\sin\theta} r \, dz = r[-2r\sin\theta] = -2r^2\sin\theta
   \]
2. **Integrate with respect to \( r \):**
   \[
   \int_{r=0}^{5\cos\theta} -2r^2\sin\theta \, dr = -2\sin\theta \int_{0}^{5\cos\theta} r^2 \, dr = -2\sin\theta \left[\frac{r^3}{3}\right]_{0}^{5\cos\theta} = -2\sin\theta \cdot \frac{(5\cos\theta)^3}{3} = -\frac{250\cos^3\theta\sin\theta}{3}
   \]
3. **Integrate with respect to \( \theta \):**
   \[
   V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\frac{250\cos^3\theta\sin\theta}{3} \, d\theta
   \]
   Let \( u = \cos\theta \), then \( du = -\sin\theta \, d\theta \), and the limits change from \( \cos\left(-\frac{\pi}{2}\right) = 0 \) to \( \cos\left(\frac{\pi}{2}\right) = 0 \) through 1 at \( \theta = 0 \).

   The integral simplifies to zero because it is an odd function over a symmetric interval.

### Conclusion:
The volume of the solid is \( \boxed{0} \). This result occurs because the region above and below the \( xy \)-plane cancels out.

Would you like further details, or do you have any questions?

### Related Questions:
1. How do cylindrical coordinates work in integration?
2. What is the significance of a function being odd over a symmetric interval?
3. How can symmetry simplify volume calculations?
4. What are the bounds for polar integrals involving cardioids?
5. How can we use transformations to evaluate triple integrals?

### Tip:
Always check the symmetry of a region before integrating, as it can simplify or even eliminate the need for computation.

Learn how to calculate the volume of a solid bounded by z = -2y, the xy-plane, and r = 5cos(theta) using cylindrical coordinates and triple integrals. Understand the concept of symmetry in integrals and the application of cylindrical coordinate transformations. Suitable for advanced college-level mathematics.

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