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Sagot :
To show that the volume of a sphere with radius [tex]\( r \)[/tex] is [tex]\( V = \frac{4}{3} \pi r^3 \)[/tex], we will use the method of integrating small volume elements.
### Step-by-Step Solution:
1. Surface Area of Sphere:
- The surface area [tex]\( A \)[/tex] of a sphere with radius [tex]\( r \)[/tex] is given by:
[tex]\[ A = 4 \pi r^2 \][/tex]
2. Small Volume Element:
- Imagine dividing the sphere into an infinite number of thin spherical shells, each with a very small thickness [tex]\( \Delta x \)[/tex].
- The radius of each small shell is [tex]\( x \)[/tex].
3. Volume of a Thin Spherical Shell:
- The volume [tex]\( \Delta V \)[/tex] of a thin spherical shell of radius [tex]\( x \)[/tex] and thickness [tex]\( \Delta x \)[/tex] is approximately the surface area at radius [tex]\( x \)[/tex] multiplied by the thickness.
- Thus, the volume [tex]\( \Delta V \)[/tex] of the shell is:
[tex]\[ \Delta V \approx 4 \pi x^2 \Delta x \][/tex]
4. Integration to Find the Total Volume:
- To find the total volume of the sphere, we sum the volumes of all the thin shells from [tex]\( x = 0 \)[/tex] to [tex]\( x = r \)[/tex].
- This summation is represented by the integral:
[tex]\[ V = \int_{0}^{r} 4 \pi x^2 \, dx \][/tex]
5. Evaluating the Integral:
- We need to evaluate the integral:
[tex]\[ \int_{0}^{r} 4 \pi x^2 \, dx \][/tex]
- First, factor out the constants:
[tex]\[ V = 4 \pi \int_{0}^{r} x^2 \, dx \][/tex]
6. Integral of [tex]\( x^2 \)[/tex]:
- Use the power rule for integration:
[tex]\[ \int x^2 \, dx = \frac{x^3}{3} \][/tex]
- Applying this to our integral, we get:
[tex]\[ V = 4 \pi \left[ \frac{x^3}{3} \right]_{0}^{r} \][/tex]
- Evaluate the definite integral:
[tex]\[ V = 4 \pi \left( \frac{r^3}{3} - \frac{0^3}{3} \right) \][/tex]
[tex]\[ V = 4 \pi \left( \frac{r^3}{3} \right) \][/tex]
[tex]\[ V = \frac{4}{3} \pi r^3 \][/tex]
### Conclusion:
Thus, the volume of a sphere with radius [tex]\( r \)[/tex] is indeed [tex]\( V = \frac{4}{3} \pi r^3 \)[/tex]. This derivation aligns perfectly with the formula for the volume of a sphere.
### Step-by-Step Solution:
1. Surface Area of Sphere:
- The surface area [tex]\( A \)[/tex] of a sphere with radius [tex]\( r \)[/tex] is given by:
[tex]\[ A = 4 \pi r^2 \][/tex]
2. Small Volume Element:
- Imagine dividing the sphere into an infinite number of thin spherical shells, each with a very small thickness [tex]\( \Delta x \)[/tex].
- The radius of each small shell is [tex]\( x \)[/tex].
3. Volume of a Thin Spherical Shell:
- The volume [tex]\( \Delta V \)[/tex] of a thin spherical shell of radius [tex]\( x \)[/tex] and thickness [tex]\( \Delta x \)[/tex] is approximately the surface area at radius [tex]\( x \)[/tex] multiplied by the thickness.
- Thus, the volume [tex]\( \Delta V \)[/tex] of the shell is:
[tex]\[ \Delta V \approx 4 \pi x^2 \Delta x \][/tex]
4. Integration to Find the Total Volume:
- To find the total volume of the sphere, we sum the volumes of all the thin shells from [tex]\( x = 0 \)[/tex] to [tex]\( x = r \)[/tex].
- This summation is represented by the integral:
[tex]\[ V = \int_{0}^{r} 4 \pi x^2 \, dx \][/tex]
5. Evaluating the Integral:
- We need to evaluate the integral:
[tex]\[ \int_{0}^{r} 4 \pi x^2 \, dx \][/tex]
- First, factor out the constants:
[tex]\[ V = 4 \pi \int_{0}^{r} x^2 \, dx \][/tex]
6. Integral of [tex]\( x^2 \)[/tex]:
- Use the power rule for integration:
[tex]\[ \int x^2 \, dx = \frac{x^3}{3} \][/tex]
- Applying this to our integral, we get:
[tex]\[ V = 4 \pi \left[ \frac{x^3}{3} \right]_{0}^{r} \][/tex]
- Evaluate the definite integral:
[tex]\[ V = 4 \pi \left( \frac{r^3}{3} - \frac{0^3}{3} \right) \][/tex]
[tex]\[ V = 4 \pi \left( \frac{r^3}{3} \right) \][/tex]
[tex]\[ V = \frac{4}{3} \pi r^3 \][/tex]
### Conclusion:
Thus, the volume of a sphere with radius [tex]\( r \)[/tex] is indeed [tex]\( V = \frac{4}{3} \pi r^3 \)[/tex]. This derivation aligns perfectly with the formula for the volume of a sphere.
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