Find the volume of the solid generated by revolving the region bounded by the parabola $$ y=\frac{x^{2}}{16} $$ and the line $$ y=1 $$ about the following lines. $$ \begin{array}{l} ext { a. The line } y=1 \ ext { b. The line } y=2 \ ext { c. The line } y=-1 \ ext { (Type an exact answer, using } \pi ext { ras needed.) }\end{array} $$

Question

Find the volume of the solid generated by revolving the region bounded by the parabola $$ y=\frac{x^{2}}{16} $$ and the line $$ y=1 $$ about the following lines. $$ \begin{array}{l}	ext { a. The line } y=1 \ 	ext { b. The line } y=2 \ 	ext { c. The line } y=-1 \ 	ext { (Type an exact answer, using } \pi 	ext { ras needed.) }\end{array} $$

UpStudy AI · Accepted Answer

To find the volume of the solid generated by revolving the region bounded by the parabola $$ y = \frac{x^2}{16} $$ and the line $$ y = 1 $$ about the specified lines, we will use the method of washers.

### Step 1: Find the points of intersection
First, we need to find the points where the parabola intersects the line $$ y = 1 $$:

$$
\frac{x^2}{16} = 1
$$

Multiplying both sides by 16 gives:

$$
x^2 = 16
$$

Taking the square root of both sides, we find:

$$
x = 4 \quad \text{and} \quad x = -4
$$

Thus, the region of interest is bounded by $$ x = -4 $$ and $$ x = 4 $$.

### Step 2: Volume when revolving about $$ y = 1 $$
When revolving around the line $$ y = 1 $$, the outer radius $$ R $$ is the distance from $$ y = 1 $$ to the parabola, and the inner radius $$ r $$ is the distance from $$ y = 1 $$ to the line itself (which is 0 since it is the same line).

The outer radius $$ R $$ is given by:

$$
R = 1 - \frac{x^2}{16}
$$

The volume $$ V $$ is given by the integral:

$$
V = \pi \int_{-4}^{4} \left( R^2 - r^2 \right) \, dx = \pi \int_{-4}^{4} \left( \left(1 - \frac{x^2}{16}\right)^2 - 0^2 \right) \, dx
$$

### Step 3: Volume when revolving about $$ y = 2 $$
When revolving around the line $$ y = 2 $$, the outer radius $$ R $$ is the distance from $$ y = 2 $$ to the parabola, and the inner radius $$ r $$ is the distance from $$ y = 2 $$ to the line $$ y = 1 $$.

The outer radius $$ R $$ is given by:

$$
R = 2 - \frac{x^2}{16}
$$

The inner radius $$ r $$ is:

$$
r = 2 - 1 = 1
$$

The volume $$ V $$ is given by the integral:

$$
V = \pi \int_{-4}^{4} \left( R^2 - r^2 \right) \, dx = \pi \int_{-4}^{4} \left( \left(2 - \frac{x^2}{16}\right)^2 - 1^2 \right) \, dx
$$

### Step 4: Volume when revolving about $$ y = -1 $$
When revolving around the line $$ y = -1 $$, the outer radius $$ R $$ is the distance from $$ y = -1 $$ to the line $$ y = 1 $$, and the inner radius $$ r $$ is the distance from $$ y = -1 $$ to the parabola.

The outer radius $$ R $$ is:

$$
R = 1 - (-1) = 2
$$

The inner radius $$ r $$ is:

$$
r = 1 - (-1) = 2 + \frac{x^2}{16}
$$

The volume $$ V $$ is given by the integral:

$$
V = \pi \int_{-4}^{4} \left( R^2 - r^2 \right) \, dx = \pi \int_{-4}^{4} \left( 2^2 - \left(1 + \frac{x^2}{16}\right)^2 \right) \, dx
$$

Now, let's compute these integrals.

### Step 5: Calculate the volumes
We will compute the three volumes using the integrals derived above.

1. For $$ y = 1 $$:
   $$
   V_1 = \pi \int_{-4}^{4} \left( 1 - \frac{x^2}{16} \right)^2 \, dx
   $$

2. For $$ y = 2 $$:
   $$
   V_2 = \pi \int_{-4}^{4} \left( \left(2 - \frac{x^2}{16}\right)^2 - 1^2 \right) \, dx
   $$

3. For $$ y = -1 $$:
   $$
   V_3 = \pi \int_{-4}^{4} \left( 2^2 - \left(1 + \frac{x^2}{16}\right)^2 \right) \, dx
   $$

Now, I will calculate these integrals.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{-4}^{4} \left(2^{2}-\left(1+\frac{x^{2}}{16}\right)^{2}\right) dx$$
- step1: Subtract the terms:
  $$\int_{-4}^{4} \left(3-\frac{1}{8}x^{2}-\frac{1}{256}x^{4}\right) dx$$
- step2: Evaluate the integral:
  $$\int \left(3-\frac{1}{8}x^{2}-\frac{1}{256}x^{4}\right) dx$$
- step3: Use properties of integrals:
  $$\int 3 dx+\int -\frac{1}{8}x^{2} dx+\int -\frac{1}{256}x^{4} dx$$
- step4: Evaluate the integral:
  $$3x+\int -\frac{1}{8}x^{2} dx+\int -\frac{1}{256}x^{4} dx$$
- step5: Evaluate the integral:
  $$3x-\frac{x^{3}}{24}+\int -\frac{1}{256}x^{4} dx$$
- step6: Evaluate the integral:
  $$3x-\frac{x^{3}}{24}-\frac{x^{5}}{1280}$$
- step7: Return the limits:
  $$\left(3x-\frac{x^{3}}{24}-\frac{x^{5}}{1280}\right)\bigg |_{-4}^{4}$$
- step8: Calculate the value:
  $$\frac{256}{15}$$
Calculate or simplify the expression $$ \int_{-4}^{4} \left( 1 - \frac{x^2}{16} \right)^2 \, dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{-4}^{4} \left(1-\frac{x^{2}}{16}\right)^{2} dx$$
- step1: Evaluate the integral:
  $$\int \left(1-\frac{x^{2}}{16}\right)^{2} dx$$
- step2: Calculate:
  $$\int \left(1-\frac{x^{2}}{16}-\frac{x^{2}}{16}+\frac{x^{4}}{256}\right) dx$$
- step3: Use properties of integrals:
  $$\int 1 dx+\int -\frac{x^{2}}{16} dx+\int -\frac{x^{2}}{16} dx+\int \frac{x^{4}}{256} dx$$
- step4: Evaluate the integral:
  $$x+\int -\frac{x^{2}}{16} dx+\int -\frac{x^{2}}{16} dx+\int \frac{x^{4}}{256} dx$$
- step5: Evaluate the integral:
  $$x-\frac{x^{3}}{48}+\int -\frac{x^{2}}{16} dx+\int \frac{x^{4}}{256} dx$$
- step6: Evaluate the integral:
  $$x-\frac{x^{3}}{48}-\frac{x^{3}}{48}+\int \frac{x^{4}}{256} dx$$
- step7: Evaluate the integral:
  $$x-\frac{x^{3}}{48}-\frac{x^{3}}{48}+\frac{x^{5}}{1280}$$
- step8: Return the limits:
  $$\left(x-\frac{x^{3}}{48}-\frac{x^{3}}{48}+\frac{x^{5}}{1280}\right)\bigg |_{-4}^{4}$$
- step9: Calculate the value:
  $$\frac{64}{15}$$
Calculate or simplify the expression $$ \int_{-4}^{4} \left( \left(2 - \frac{x^2}{16}\right)^2 - 1^2 \right) \, dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{-4}^{4} \left(\left(2-\frac{x^{2}}{16}\right)^{2}-1^{2}\right) dx$$
- step1: Evaluate the power:
  $$\int_{-4}^{4} \left(\left(2-\frac{x^{2}}{16}\right)^{2}-1\right) dx$$
- step2: Subtract the terms:
  $$\int_{-4}^{4} \left(3-\frac{1}{4}x^{2}+\frac{1}{256}x^{4}\right) dx$$
- step3: Evaluate the integral:
  $$\int \left(3-\frac{1}{4}x^{2}+\frac{1}{256}x^{4}\right) dx$$
- step4: Use properties of integrals:
  $$\int 3 dx+\int -\frac{1}{4}x^{2} dx+\int \frac{1}{256}x^{4} dx$$
- step5: Evaluate the integral:
  $$3x+\int -\frac{1}{4}x^{2} dx+\int \frac{1}{256}x^{4} dx$$
- step6: Evaluate the integral:
  $$3x-\frac{x^{3}}{12}+\int \frac{1}{256}x^{4} dx$$
- step7: Evaluate the integral:
  $$3x-\frac{x^{3}}{12}+\frac{x^{5}}{1280}$$
- step8: Return the limits:
  $$\left(3x-\frac{x^{3}}{12}+\frac{x^{5}}{1280}\right)\bigg |_{-4}^{4}$$
- step9: Calculate the value:
  $$\frac{224}{15}$$
The volumes of the solid generated by revolving the region bounded by the parabola $$ y = \frac{x^2}{16} $$ and the line $$ y = 1 $$ about the specified lines are as follows:

### a. Volume when revolving about the line $$ y = 1 $$
$$
V_1 = \pi \cdot \frac{64}{15} = \frac{64\pi}{15}
$$

### b. Volume when revolving about the line $$ y = 2 $$
$$
V_2 = \pi \cdot \frac{224}{15} = \frac{224\pi}{15}
$$

### c. Volume when revolving about the line $$ y = -1 $$
$$
V_3 = \pi \cdot \frac{256}{15} = \frac{256\pi}{15}
$$

Thus, the exact volumes are:
- $$ V_1 = \frac{64\pi}{15} $$
- $$ V_2 = \frac{224\pi}{15} $$
- $$ V_3 = \frac{256\pi}{15} $$
The volumes of the solid generated by revolving the region bounded by the parabola $$ y = \frac{x^2}{16} $$ and the line $$ y = 1 $$ about the specified lines are: - **a.** $$ \frac{64\pi}{15} $$ - **b.** $$ \frac{224\pi}{15} $$ - **c.** $$ \frac{256\pi}{15} $$

Find the volume of the solid generated by revolving the region bounded by the parabola and the line
about the following lines.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Step 1: Find the points of intersection

Step 2: Volume when revolving about

Step 3: Volume when revolving about

Step 4: Volume when revolving about

Step 5: Calculate the volumes

a. Volume when revolving about the line

b. Volume when revolving about the line

c. Volume when revolving about the line

Beyond the Answer

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Find the volume of the solid generated by revolving the region bounded by the parabola and the line about the following lines.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Step 1: Find the points of intersection

Step 2: Volume when revolving about

Step 3: Volume when revolving about

Step 4: Volume when revolving about

Step 5: Calculate the volumes

a. Volume when revolving about the line

b. Volume when revolving about the line

c. Volume when revolving about the line

Beyond the Answer

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Latest Calculus Questions

Find the volume of the solid generated by revolving the region bounded by the parabola and the line
about the following lines.