Question 1175395
Let's break down this problem step-by-step.

**1. Define the Ellipse**

* The bowl is a lower half of a spheroid, so its vertical cross-sections are semi-ellipses.
* The bowl is 10 inches wide at the opening, so the major axis of the ellipse is 10 inches. Thus, the semi-major axis (a) is 5 inches.
* The bowl is 10 inches deep at the center, so the semi-minor axis (b) is 10 inches.
* The equation of the ellipse, centered at the origin, is:
    * x²/a² + y²/b² = 1
    * x²/5² + y²/10² = 1
    * x²/25 + y²/100 = 1

**2. Relate the Circular Cover to the Ellipse**

* The circular cover has a diameter of 9 inches, so its radius (r) is 4.5 inches.
* We want to find the depth (y) at which the cover fits into the bowl.
* When the cover fits into the bowl, the x-coordinate of the ellipse is equal to the radius of the circular cover.
* Therefore, x = 4.5 inches.

**3. Solve for the Depth (y)**

* Substitute x = 4.5 into the ellipse equation:
    * (4.5)²/25 + y²/100 = 1
    * 20.25/25 + y²/100 = 1
    * 0.81 + y²/100 = 1
    * y²/100 = 1 - 0.81
    * y²/100 = 0.19
    * y² = 19
    * y = √19
    * y ≈ 4.3589 inches

**4. Calculate the Depth from the Top**

* The total depth of the bowl is 10 inches.
* The y-coordinate we found (√19) is the distance from the bottom of the bowl.
* The depth from the top of the bowl is:
    * 10 - √19 ≈ 10 - 4.3589 ≈ 5.6411 inches

**Answer:**

The circular cover with a diameter of 9 inches goes approximately 5.64 inches deep into the bowl.