Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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.
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
