Find the volume generated by revolving the area under the curve๐ฆ = ๐๐ฅ, from ๐ฅ = 0 to ๐ฅ =1,
about the line ๐ฅ = 1.
using shells of thickness dx, we have
v = โซ[0,1] 2ฯrh dx
where r=1-x and h=y
= โซ[0,1] ฯ(1-x)^2*e^x dx = (2e-5)ฯ
To find the volume generated by revolving the area under the curve ๐ฆ = ๐๐ฅ from ๐ฅ = 0 to ๐ฅ = 1 about the line ๐ฅ = 1, we will use the method of cylindrical shells.
The formula to calculate the volume using cylindrical shells is:
๐ = โซ (2๐๐ฆ)(๐ฅ)(๐๐ฅ)
In this case, ๐ฆ = ๐๐ฅ, so we will substitute ๐ฆ = ๐๐ฅ in the formula:
๐ = โซ (2๐๐๐ฅ)(๐ฅ)(๐๐ฅ)
To integrate this expression, we need to find the limits of integration. In this case, the limits are from ๐ฅ = 0 to ๐ฅ = 1, as mentioned.
๐ = โซโยน (2๐๐๐ฅ)(๐ฅ)(๐๐ฅ)
Now we can proceed with the integration:
๐ = 2๐ โซโยน ๐๐ฅ๐ฅยฒ ๐๐ฅ
To integrate ๐๐ฅ๐ฅยฒ, we can use integration by parts. Let's define ๐ข and ๐ฃ as:
๐ข = ๐ฅยฒ โ ๐๐ข = 2๐ฅ๐๐ฅ
๐๐ฃ = ๐๐ฅ๐๐ฅ โ ๐ฃ = ๐๐ฅ
Using the integration by parts formula:
โซ ๐ข๐๐ฃ = ๐ข๐ฃ - โซ ๐ฃ๐๐ข
Substituting ๐ข, ๐๐ข, ๐ฃ, and ๐๐ฃ:
โซ ๐ฅยฒ๐๐ฅ๐๐ฅ = ๐ฅยฒ๐๐ฅ - โซ (2๐ฅ)(๐๐ฅ)๐๐ฅ
Simplifying this equation:
โซ ๐ฅยฒ๐๐ฅ๐๐ฅ = ๐ฅยฒ๐๐ฅ - 2 โซ ๐ฅ๐๐ฅ๐๐ฅ
Now, we need to evaluate the integral of ๐ฅ๐๐ฅ. This integral can be solved using integration by parts again.
Let's define ๐ข and ๐ฃ as:
๐ข = ๐ฅ โ ๐๐ข = ๐๐ฅ
๐๐ฃ = ๐๐ฅ โ ๐ฃ = ๐๐ฅ
Using the integration by parts formula:
โซ ๐ข๐๐ฃ = ๐ข๐ฃ - โซ ๐ฃ๐๐ข
Substituting ๐ข, ๐๐ข, ๐ฃ, and ๐๐ฃ:
โซ ๐ฅ๐๐ฅ๐๐ฅ = ๐ฅ๐๐ฅ - โซ ๐๐ฅ๐๐ฅ
We can evaluate this integral:
โซ ๐ฅ๐๐ฅ๐๐ฅ = ๐ฅ๐๐ฅ - ๐๐ฅ + ๐ถ
Substituting this result back into our previous equation:
โซ ๐ฅยฒ๐๐ฅ๐๐ฅ = ๐ฅยฒ๐๐ฅ - 2(๐ฅ๐๐ฅ - ๐๐ฅ + ๐ถ)
Simplifying:
โซ ๐ฅยฒ๐๐ฅ๐๐ฅ = ๐ฅยฒ๐๐ฅ - 2๐ฅ๐๐ฅ + 2๐๐ฅ + ๐ถ
Now, we can substitute this back into the integral we were calculating initially:
๐ = 2๐ โซโยน ๐๐ฅ๐ฅยฒ ๐๐ฅ
= 2๐ [๐ฅยฒ๐๐ฅ - 2๐ฅ๐๐ฅ + 2๐๐ฅ] ๐๐๐ข๐๐ from ๐ฅ = 0 to ๐ฅ = 1
Evaluating this integral:
๐ = 2๐ [(1ยฒ๐ยน - 2(1)๐ยน + 2๐ยน) - (0ยฒ๐โฐ - 2(0)๐โฐ + 2๐โฐ)]
Simplifying:
๐ = 2๐ [๐ - 2๐ + 2๐ - 1]
= 2๐ (๐ - 1)
So, the volume generated by revolving the area under the curve ๐ฆ = ๐๐ฅ, from ๐ฅ = 0 to ๐ฅ = 1, about the line ๐ฅ = 1 is 2๐(๐ - 1) cubic units.
To find the volume generated by revolving the area under the curve ๐ฆ = ๐๐ฅ from ๐ฅ = 0 to ๐ฅ = 1 about the line ๐ฅ = 1, we can use the disk method.
The disk method involves adding up the volumes of infinitely many thin disks, each with a radius determined by the curve, and then integrating to find the total volume.
To start, let's visualize the situation. The curve ๐ฆ = ๐๐ฅ looks like an exponential curve that starts at (0, 1) and increases rapidly. We are revolving this curve about the vertical line ๐ฅ = 1.
To find the radius of each disk, we need to consider that each point on the curve is a distance of 1 unit away from the line ๐ฅ = 1. So, the radius of each disk is the horizontal distance from the point on the curve to the line ๐ฅ = 1, which is 1 - ๐ฅ.
To find the volume of each disk, we can use the formula for the volume of a cylinder: V = ฯ * r^2 * h, where r is the radius and h is the height. In this case, the height of each disk is given by the curve ๐ฆ = ๐๐ฅ, and the radius is given by 1 - ๐ฅ.
Now, to find the total volume, we need to integrate the formula for the volume of each disk from ๐ฅ = 0 to ๐ฅ = 1:
V = โซ[0 to 1] ฯ * (1 - ๐ฅ)^2 * ๐๐ฅ ๐๐ฅ
To evaluate this integral, we can use integration techniques such as integration by parts or substitution. After evaluating the integral, we will have the value of the volume generated by revolving the area under the curve ๐ฆ = ๐๐ฅ from ๐ฅ = 0 to ๐ฅ = 1 about the line ๐ฅ = 1.