Question
Let H be the hemisphere (x^2 + y^2 + z^2) = 45, z >= 0, and suppose f is a continuous function with f(5,2,4)=8, f(-5,2,4)=10, and f(-5,-2,4)=13. By dividing H into four patches, estimate the value below, round to the nearest whole number:
Double integral bound by H of f(x,y,z) dS
Double integral bound by H of f(x,y,z) dS
Answers
GPT-4o mini
To estimate the double integral of the function \( f \) over the surface of the hemisphere \( H \), we will divide the hemisphere into four patches and approximate the integrals over these patches.
### Step 1: Understand the Geometry
The hemisphere \( H \) is defined by the equation:
\[
x^2 + y^2 + z^2 = 45, \quad z \geq 0
\]
The radius of the hemisphere is \( \sqrt{45} = 3\sqrt{5} \).
### Step 2: Points in the Hemisphere
Given points provided are:
- \( (5, 2, 4) \)
- \( (-5, 2, 4) \)
- \( (-5, -2, 4) \)
We also note that \( f \) is given at these points.
### Step 3: Divide Hemisphere into Four Patches
To create four patches, we can split the circular base of the hemisphere in the \(xy\)-plane into four quadrants:
1. **First Quadrant**: \( 0 \leq x \leq 5, 0 \leq y \leq 5 \)
2. **Second Quadrant**: \( -5 \leq x < 0, 0 \leq y \leq 5 \)
3. **Third Quadrant**: \( -5 \leq x < 0, -5 < y < 0 \)
4. **Fourth Quadrant**: \( 0 \leq x < 5, -5 < y < 0 \)
However, since the points \( (5, 2, 4) \), \( (-5, 2, 4) \), and \( (-5, -2, 4) \) fall into the second and third quadrants, we focus on those areas.
### Step 4: Calculate Area of Each Patch
For approximating the double integral, we will compute the average of \( f \) and multiply by the area of the surface patch. The area element \( dS \) on the hemisphere can be approximated by using spherical coordinates or geometry, but we will focus on averages and estimation.
For simplicity and given the symmetric properties of the function concerning \( y \) in the hemisphere, we can reduce calculation complexity by averaging the values of \( f \):
1. For the second quadrant (patch around \( (-5, 2, 4) \)):
- Approximate \( f \) by taking the value at \( (-5, 2, 4) = 10 \).
2. For the third quadrant (patch around \( (-5, -2, 4) \)):
- Use the point \( (-5, -2, 4) = 13 \).
### Step 5: Average Values
For estimation, since no corresponding fourth quadrant point was derived:
Averaging values (exclude the isolated point, use points accordingly):
- Use \( f(-5, 2, 4) \) and \( f(-5, -2, 4) \):
\[
\text{Average value of } f \approx \frac{10 + 13}{2} = 11.5
\]
### Step 6: Estimate Surface Area of Hemisphere
The total area \( A \) of the hemisphere can be calculated as:
\[
A = \frac{1}{2} \times 4\pi r^2 = 2\pi(45) = 90\pi
\]
### Step 7: Combine for Integral Result
Thus the double integral estimate becomes:
\[
\iint_{H} f(x,y,z) \, dS \approx \text{average value} \times \text{area}
\]
Finally,
\[
\iint_{H} f \, dS \approx 11.5 \times 90\pi
\]
To compute this:
Using \( \pi \approx 3.14 \):
\[
= 11.5 \times 90 \times 3.14 \approx 11.5 \times 282.6 \approx 3250.9
\]
Rounding gives:
\[
\text{Final estimate } \approx 3251
\]
### Conclusion
The estimated double integral bound by the hemisphere of \( f(x,y,z) \) rounded to the nearest whole number is:
\[
\boxed{3251}
\]
### Step 1: Understand the Geometry
The hemisphere \( H \) is defined by the equation:
\[
x^2 + y^2 + z^2 = 45, \quad z \geq 0
\]
The radius of the hemisphere is \( \sqrt{45} = 3\sqrt{5} \).
### Step 2: Points in the Hemisphere
Given points provided are:
- \( (5, 2, 4) \)
- \( (-5, 2, 4) \)
- \( (-5, -2, 4) \)
We also note that \( f \) is given at these points.
### Step 3: Divide Hemisphere into Four Patches
To create four patches, we can split the circular base of the hemisphere in the \(xy\)-plane into four quadrants:
1. **First Quadrant**: \( 0 \leq x \leq 5, 0 \leq y \leq 5 \)
2. **Second Quadrant**: \( -5 \leq x < 0, 0 \leq y \leq 5 \)
3. **Third Quadrant**: \( -5 \leq x < 0, -5 < y < 0 \)
4. **Fourth Quadrant**: \( 0 \leq x < 5, -5 < y < 0 \)
However, since the points \( (5, 2, 4) \), \( (-5, 2, 4) \), and \( (-5, -2, 4) \) fall into the second and third quadrants, we focus on those areas.
### Step 4: Calculate Area of Each Patch
For approximating the double integral, we will compute the average of \( f \) and multiply by the area of the surface patch. The area element \( dS \) on the hemisphere can be approximated by using spherical coordinates or geometry, but we will focus on averages and estimation.
For simplicity and given the symmetric properties of the function concerning \( y \) in the hemisphere, we can reduce calculation complexity by averaging the values of \( f \):
1. For the second quadrant (patch around \( (-5, 2, 4) \)):
- Approximate \( f \) by taking the value at \( (-5, 2, 4) = 10 \).
2. For the third quadrant (patch around \( (-5, -2, 4) \)):
- Use the point \( (-5, -2, 4) = 13 \).
### Step 5: Average Values
For estimation, since no corresponding fourth quadrant point was derived:
Averaging values (exclude the isolated point, use points accordingly):
- Use \( f(-5, 2, 4) \) and \( f(-5, -2, 4) \):
\[
\text{Average value of } f \approx \frac{10 + 13}{2} = 11.5
\]
### Step 6: Estimate Surface Area of Hemisphere
The total area \( A \) of the hemisphere can be calculated as:
\[
A = \frac{1}{2} \times 4\pi r^2 = 2\pi(45) = 90\pi
\]
### Step 7: Combine for Integral Result
Thus the double integral estimate becomes:
\[
\iint_{H} f(x,y,z) \, dS \approx \text{average value} \times \text{area}
\]
Finally,
\[
\iint_{H} f \, dS \approx 11.5 \times 90\pi
\]
To compute this:
Using \( \pi \approx 3.14 \):
\[
= 11.5 \times 90 \times 3.14 \approx 11.5 \times 282.6 \approx 3250.9
\]
Rounding gives:
\[
\text{Final estimate } \approx 3251
\]
### Conclusion
The estimated double integral bound by the hemisphere of \( f(x,y,z) \) rounded to the nearest whole number is:
\[
\boxed{3251}
\]