Find the volume generated by revolving the area under the curve𝑦 = 𝑒𝑥, from 𝑥 = 0 to 𝑥 =1,

Question

Find the volume generated by revolving the area under the curve𝑦 = 𝑒𝑥, from 𝑥 = 0 to 𝑥 =1,

about the line 𝑥 = 1.

Answer 1

using shells of thickness dx, we have

v = ∫[0,1] 2πrh dx
where r=1-x and h=y
= ∫[0,1] π(1-x)^2*e^x dx = (2e-5)π

Answer 2

To find the volume generated by revolving the area under the curve 𝑦 = 𝑒𝑥 from 𝑥 = 0 to 𝑥 = 1 about the line 𝑥 = 1, we will use the method of cylindrical shells.

The formula to calculate the volume using cylindrical shells is:

𝑉 = ∫ (2𝜋𝑦)(𝑥)(𝑑𝑥)

In this case, 𝑦 = 𝑒𝑥, so we will substitute 𝑦 = 𝑒𝑥 in the formula:

𝑉 = ∫ (2𝜋𝑒𝑥)(𝑥)(𝑑𝑥)

To integrate this expression, we need to find the limits of integration. In this case, the limits are from 𝑥 = 0 to 𝑥 = 1, as mentioned.

𝑉 = ∫₀¹ (2𝜋𝑒𝑥)(𝑥)(𝑑𝑥)

Now we can proceed with the integration:

𝑉 = 2𝜋 ∫₀¹ 𝑒𝑥𝑥² 𝑑𝑥

To integrate 𝑒𝑥𝑥², we can use integration by parts. Let's define 𝑢 and 𝑣 as:

𝑢 = 𝑥² → 𝑑𝑢 = 2𝑥𝑑𝑥
𝑑𝑣 = 𝑒𝑥𝑑𝑥 → 𝑣 = 𝑒𝑥

Using the integration by parts formula:

∫ 𝑢𝑑𝑣 = 𝑢𝑣 - ∫ 𝑣𝑑𝑢

Substituting 𝑢, 𝑑𝑢, 𝑣, and 𝑑𝑣:

∫ 𝑥²𝑒𝑥𝑑𝑥 = 𝑥²𝑒𝑥 - ∫ (2𝑥)(𝑒𝑥)𝑑𝑥

Simplifying this equation:

∫ 𝑥²𝑒𝑥𝑑𝑥 = 𝑥²𝑒𝑥 - 2 ∫ 𝑥𝑒𝑥𝑑𝑥

Now, we need to evaluate the integral of 𝑥𝑒𝑥. This integral can be solved using integration by parts again.

Let's define 𝑢 and 𝑣 as:

𝑢 = 𝑥 → 𝑑𝑢 = 𝑑𝑥
𝑑𝑣 = 𝑒𝑥 → 𝑣 = 𝑒𝑥

Using the integration by parts formula:

∫ 𝑢𝑑𝑣 = 𝑢𝑣 - ∫ 𝑣𝑑𝑢

Substituting 𝑢, 𝑑𝑢, 𝑣, and 𝑑𝑣:

∫ 𝑥𝑒𝑥𝑑𝑥 = 𝑥𝑒𝑥 - ∫ 𝑒𝑥𝑑𝑥

We can evaluate this integral:

∫ 𝑥𝑒𝑥𝑑𝑥 = 𝑥𝑒𝑥 - 𝑒𝑥 + 𝐶

Substituting this result back into our previous equation:

∫ 𝑥²𝑒𝑥𝑑𝑥 = 𝑥²𝑒𝑥 - 2(𝑥𝑒𝑥 - 𝑒𝑥 + 𝐶)

Simplifying:

∫ 𝑥²𝑒𝑥𝑑𝑥 = 𝑥²𝑒𝑥 - 2𝑥𝑒𝑥 + 2𝑒𝑥 + 𝐶

Now, we can substitute this back into the integral we were calculating initially:

𝑉 = 2𝜋 ∫₀¹ 𝑒𝑥𝑥² 𝑑𝑥
= 2𝜋 [𝑥²𝑒𝑥 - 2𝑥𝑒𝑥 + 2𝑒𝑥] 𝑏𝑜𝑢𝑛𝑑 from 𝑥 = 0 to 𝑥 = 1

Evaluating this integral:

𝑉 = 2𝜋 [(1²𝑒¹ - 2(1)𝑒¹ + 2𝑒¹) - (0²𝑒⁰ - 2(0)𝑒⁰ + 2𝑒⁰)]

Simplifying:

𝑉 = 2𝜋 [𝑒 - 2𝑒 + 2𝑒 - 1]
= 2𝜋 (𝑒 - 1)

So, the volume generated by revolving the area under the curve 𝑦 = 𝑒𝑥, from 𝑥 = 0 to 𝑥 = 1, about the line 𝑥 = 1 is 2𝜋(𝑒 - 1) cubic units.

Answer 3

To find the volume generated by revolving the area under the curve 𝑦 = 𝑒𝑥 from 𝑥 = 0 to 𝑥 = 1 about the line 𝑥 = 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 𝑦 = 𝑒𝑥 looks like an exponential curve that starts at (0, 1) and increases rapidly. We are revolving this curve about the vertical line 𝑥 = 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 𝑥 = 1. So, the radius of each disk is the horizontal distance from the point on the curve to the line 𝑥 = 1, which is 1 - 𝑥.

To find the volume of each disk, we can use the formula for the volume of a cylinder: V = π * 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 𝑦 = 𝑒𝑥, and the radius is given by 1 - 𝑥.

Now, to find the total volume, we need to integrate the formula for the volume of each disk from 𝑥 = 0 to 𝑥 = 1:

V = ∫[0 to 1] π * (1 - 𝑥)^2 * 𝑒𝑥 𝑑𝑥

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 𝑦 = 𝑒𝑥 from 𝑥 = 0 to 𝑥 = 1 about the line 𝑥 = 1.