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To find the volume of the solid formed by revolving the region enclosed by the curves [tex]\( y = \sqrt{\sqrt{x}+2} \)[/tex], [tex]\( y = 2 \)[/tex], and the line [tex]\( x = 4 \)[/tex] around the x-axis, we can use the disk method. Here’s a detailed step-by-step solution:
1. Identify the curves and bounds:
- The function given is [tex]\( y = \sqrt{\sqrt{x} + 2} \)[/tex].
- The horizontal line is [tex]\( y = 2 \)[/tex].
- The vertical line [tex]\( x = 4 \)[/tex] acts as an upper bound.
- The region starts at [tex]\( x = 0 \)[/tex] (as implied by context).
2. Set up the volume integral using the disk method:
The disk method revolves the region around the x-axis, creating disks with thickness [tex]\( dx \)[/tex]. The volume of each infinitesimally thin disk is given by the area of the disk multiplied by [tex]\( dx \)[/tex].
3. Determine the outer and inner radii:
- The outer radius [tex]\( R \)[/tex] of the disk is given by the distance from the x-axis to [tex]\( y = 2 \)[/tex]. Hence, [tex]\( R(x) = 2 \)[/tex].
- The inner radius [tex]\( r \)[/tex] of the washer is given by the distance from the x-axis to [tex]\( y = \sqrt{\sqrt{x} + 2} \)[/tex].
4. Write the volume of the solid using the integral:
The volume [tex]\( V \)[/tex] is the integral of the area of the washers from [tex]\( x = 0 \)[/tex] to [tex]\( x = 4 \)[/tex].
[tex]\[ V = \pi \int_{0}^{4} \left[ R(x)^2 - r(x)^2 \right] \, dx \][/tex]
Here, [tex]\( R(x) = 2 \)[/tex] and [tex]\( r(x) = \sqrt{\sqrt{x} + 2} \)[/tex].
5. Set up the integral:
[tex]\[ V = \pi \int_{0}^{4} \left[ 2^2 - \left( \sqrt{\sqrt{x} + 2} \right)^2 \right] \, dx \][/tex]
6. Simplify the integrand:
- [tex]\( R(x)^2 = 4 \)[/tex]
- [tex]\( r(x)^2 = \sqrt{x} + 2 \)[/tex]
Hence, the integrand becomes:
[tex]\[ 4 - (\sqrt{x} + 2) = 4 - \sqrt{x} - 2 = 2 - \sqrt{x} \][/tex]
The integral expression becomes:
[tex]\[ V = \pi \int_{0}^{4} \left( 2 - \sqrt{x} \right) \, dx \][/tex]
7. Evaluate the integral:
Compute the integral in two parts:
[tex]\[ \int_{0}^{4} 2 \, dx - \int_{0}^{4} \sqrt{x} \, dx \][/tex]
- [tex]\(\int_{0}^{4} 2 \, dx = 2x \bigg|_0^4 = 2(4) - 2(0) = 8\)[/tex]
- [tex]\(\int_{0}^{4} \sqrt{x} \, dx = \int_{0}^{4} x^{1/2} \, dx = \frac{2}{3} x^{3/2} \bigg|_0^4 = \frac{2}{3} (4^{3/2}) = \frac{2}{3} (8) = \frac{16}{3}\)[/tex]
Subtracting the two results:
[tex]\[ 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3} \][/tex]
8. Multiply by [tex]\(\pi\)[/tex]:
[tex]\[ V = \pi \left( \frac{8}{3} \right) = \frac{8\pi}{3} \][/tex]
This result is then:
[tex]\[ 8.37758040957278 \text{ (in decimal form)} \][/tex]
So, the volume of the solid formed by revolving the given region about the x-axis is approximately [tex]\( 8.37758040957278 \)[/tex] cubic units.
1. Identify the curves and bounds:
- The function given is [tex]\( y = \sqrt{\sqrt{x} + 2} \)[/tex].
- The horizontal line is [tex]\( y = 2 \)[/tex].
- The vertical line [tex]\( x = 4 \)[/tex] acts as an upper bound.
- The region starts at [tex]\( x = 0 \)[/tex] (as implied by context).
2. Set up the volume integral using the disk method:
The disk method revolves the region around the x-axis, creating disks with thickness [tex]\( dx \)[/tex]. The volume of each infinitesimally thin disk is given by the area of the disk multiplied by [tex]\( dx \)[/tex].
3. Determine the outer and inner radii:
- The outer radius [tex]\( R \)[/tex] of the disk is given by the distance from the x-axis to [tex]\( y = 2 \)[/tex]. Hence, [tex]\( R(x) = 2 \)[/tex].
- The inner radius [tex]\( r \)[/tex] of the washer is given by the distance from the x-axis to [tex]\( y = \sqrt{\sqrt{x} + 2} \)[/tex].
4. Write the volume of the solid using the integral:
The volume [tex]\( V \)[/tex] is the integral of the area of the washers from [tex]\( x = 0 \)[/tex] to [tex]\( x = 4 \)[/tex].
[tex]\[ V = \pi \int_{0}^{4} \left[ R(x)^2 - r(x)^2 \right] \, dx \][/tex]
Here, [tex]\( R(x) = 2 \)[/tex] and [tex]\( r(x) = \sqrt{\sqrt{x} + 2} \)[/tex].
5. Set up the integral:
[tex]\[ V = \pi \int_{0}^{4} \left[ 2^2 - \left( \sqrt{\sqrt{x} + 2} \right)^2 \right] \, dx \][/tex]
6. Simplify the integrand:
- [tex]\( R(x)^2 = 4 \)[/tex]
- [tex]\( r(x)^2 = \sqrt{x} + 2 \)[/tex]
Hence, the integrand becomes:
[tex]\[ 4 - (\sqrt{x} + 2) = 4 - \sqrt{x} - 2 = 2 - \sqrt{x} \][/tex]
The integral expression becomes:
[tex]\[ V = \pi \int_{0}^{4} \left( 2 - \sqrt{x} \right) \, dx \][/tex]
7. Evaluate the integral:
Compute the integral in two parts:
[tex]\[ \int_{0}^{4} 2 \, dx - \int_{0}^{4} \sqrt{x} \, dx \][/tex]
- [tex]\(\int_{0}^{4} 2 \, dx = 2x \bigg|_0^4 = 2(4) - 2(0) = 8\)[/tex]
- [tex]\(\int_{0}^{4} \sqrt{x} \, dx = \int_{0}^{4} x^{1/2} \, dx = \frac{2}{3} x^{3/2} \bigg|_0^4 = \frac{2}{3} (4^{3/2}) = \frac{2}{3} (8) = \frac{16}{3}\)[/tex]
Subtracting the two results:
[tex]\[ 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3} \][/tex]
8. Multiply by [tex]\(\pi\)[/tex]:
[tex]\[ V = \pi \left( \frac{8}{3} \right) = \frac{8\pi}{3} \][/tex]
This result is then:
[tex]\[ 8.37758040957278 \text{ (in decimal form)} \][/tex]
So, the volume of the solid formed by revolving the given region about the x-axis is approximately [tex]\( 8.37758040957278 \)[/tex] cubic units.
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