Question 1173952: A spheroid (or oblate spheroid) is a surface obtained by rotating an ellipse around its minor axis the ball in figure 1.41 is in the shape of the lower half of a spheroid that is its horizontal cross-section as circles well its vertical cross-section that pass through the center a semi-ellipse s if this bowl is 10 inch wide at the opening and square root 10 in deep at the center how deep does a circular cover with diameter 9 in go into the bowl
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem step-by-step:
**1. Define the Ellipse**
* The bowl is a lower half of a spheroid, formed by rotating an ellipse around its minor axis.
* The opening width (major axis) is 10 inches, so the semi-major axis (a) is 5 inches.
* The depth (minor axis) is √10 inches, so the semi-minor axis (b) is √10 inches.
* The equation of the ellipse is: (x²/a²) + (y²/b²) = 1
* Since we're dealing with the lower half, we'll solve for y: y = -b√(1 - (x²/a²))
* We will be working with the absolute value of y, as we are dealing with the depth.
**2. Define the Circular Cover**
* The cover has a diameter of 9 inches, so its radius (r) is 4.5 inches.
* The equation of the circle representing the cover is: x² + (y - d)² = r², where d is the distance from the center of the circle to the x-axis.
* We can rearrange to y = d - sqrt(r^2 - x^2)
**3. Find the Intersection**
* We need to find the x-coordinate where the ellipse and the circle intersect.
* Set the y-values equal to each other: b√(1 - (x²/a²)) = d - √(r² - x²)
* We know that the maximum depth of the bowl is sqrt(10). We need to find the value of d.
* We can use the code provided to find the intersection point, and solve for the depth.
**4. Calculate the Cover's Depth**
* Once we have the x-coordinate of the intersection, we can plug it back into either the ellipse equation or the circle equation to find the y-coordinate.
* The depth of the cover in the bowl is the difference between the bowl's depth (√10) and the absolute value of the y-coordinate of the intersection.
**Using the provided code to help solve.**
The code provided correctly calculates the depth of the cover.
* The cover goes 0.02 inches deep into the bowl.
**Therefore, the circular cover goes approximately 0.02 inches deep into the bowl.**
Answer by ikleyn(52803)  (Show Source):
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