Question
Consider the region bounded by the graphs of , and in the first quadrant. What is the volume
solid obtained by rotating this region about the line ?

Ask by Weber Bush. in the United States

Jan 19,2025

Question

Question
Consider the region bounded by the graphs of , and in the first quadrant. What is the volume
solid obtained by rotating this region about the line ?

Ask by Weber Bush. in the United States

Jan 19,2025

UpStudy AI · Accepted Answer

To find the volume of the solid obtained by rotating the region bounded by the curves $$ y = \frac{1}{7} x^2 $$, $$ y = 26 $$, and $$ x = 10 $$ about the line $$ x = 7 $$, we can use the method of cylindrical shells.

### Step 1: Identify the region of integration
The curves intersect at the points where $$ y = \frac{1}{7} x^2 $$ and $$ y = 26 $$. To find the intersection points, we set:

$$
\frac{1}{7} x^2 = 26
$$

Multiplying both sides by 7 gives:

$$
x^2 = 182
$$

Taking the square root, we find:

$$
x = \sqrt{182} \approx 13.5
$$

However, since we are bounded by $$ x = 10 $$, we will only consider the region from $$ x = 0 $$ to $$ x = 10 $$.

### Step 2: Set up the volume integral
The volume $$ V $$ of the solid of revolution using the shell method is given by:

$$
V = 2\pi \int_{a}^{b} (radius)(height) \, dx
$$

In this case:
- The radius of the shell is the distance from the line $$ x = 7 $$ to the shell at position $$ x $$, which is $$ |7 - x| $$.
- The height of the shell is given by the difference between the upper curve $$ y = 26 $$ and the lower curve $$ y = \frac{1}{7} x^2 $$.

Thus, the height is:

$$
h = 26 - \frac{1}{7} x^2
$$

### Step 3: Define the limits of integration
We will integrate from $$ x = 0 $$ to $$ x = 10 $$.

### Step 4: Write the integral
The volume integral becomes:

$$
V = 2\pi \int_{0}^{10} (7 - x) \left(26 - \frac{1}{7} x^2\right) \, dx
$$

### Step 5: Simplify the integral
Now we can simplify the integrand:

$$
V = 2\pi \int_{0}^{10} \left( (7 - x)(26 - \frac{1}{7} x^2) \right) \, dx
$$

Expanding this gives:

$$
= 2\pi \int_{0}^{10} \left( 182 - 26x - \frac{7}{7} x^2 + \frac{1}{7} x^3 \right) \, dx
$$

$$
= 2\pi \int_{0}^{10} \left( 182 - 26x - x^2 + \frac{1}{7} x^3 \right) \, dx
$$

### Step 6: Calculate the integral
Now we will calculate the integral:

$$
V = 2\pi \left[ 182x - 13x^2 - \frac{1}{3} x^3 + \frac{1}{28} x^4 \right]_{0}^{10}
$$

Calculating at $$ x = 10 $$:

$$
= 2\pi \left[ 182(10) - 13(10^2) - \frac{1}{3}(10^3) + \frac{1}{28}(10^4) \right]
$$

Calculating each term:

- $$ 182(10) = 1820 $$
- $$ 13(10^2) = 1300 $$
- $$ \frac{1}{3}(10^3) = \frac{1000}{3} \approx 333.33 $$
- $$ \frac{1}{28}(10^4) = \frac{10000}{28} \approx 357.14 $$

Now substituting these values into the expression:

$$
= 2\pi \left[ 1820 - 1300 - 333.33 + 357.14 \right]
$$

Calculating the expression inside the brackets:

$$
= 2\pi \left[ 1820 - 1300 - 333.33 + 357.14 \right] = 2\pi \left[ 543.81 \right]
$$

### Step 7: Final volume calculation
Now we can calculate the final volume:

$$
V \approx 2\pi (543.81) \approx 1087.62\pi
$$

Thus, the volume of the solid obtained by rotating the region about the line $$ x = 7 $$ is:

$$
\boxed{1087.62\pi}
$$
The volume of the solid is $$ 1087.62\pi $$.

Question
Consider the region bounded by the graphs of , and in the first quadrant. What is the volume
solid obtained by rotating this region about the line ?

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Answer

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Question Consider the region bounded by the graphs of , and in the first quadrant. What is the volume solid obtained by rotating this region about the line ?

Upstudy AI Solution

Answer

Solution

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Question
Consider the region bounded by the graphs of , and in the first quadrant. What is the volume
solid obtained by rotating this region about the line ?