How do you find the volume of the solid generated by revolving the region bounded by the curves y = 10 / x², y = 0, x = 1, x = 5 rotated about the x-axis?

Question

Samantha Adkison · Accepted Answer

The area to be rotated should look like this one:

Emily Ammerman · Answer

undefined To find the volume of the solid generated by revolving the region bounded by the curves $ y = \frac{10}{x^2} $, $ y = 0 $, $ x = 1 $, and $ x = 5 $ rotated about the x-axis, you can use the disk method.

1. Determine the limits of integration. In this case, the region is bounded by the vertical lines $ x = 1 $ and $ x = 5 $.

2. Set up the integral for the volume using the disk method:
$$ V = \pi \int_{1}^{5} [f(x)]^2 \, dx $$
where $ f(x) $ represents the upper curve $ y = \frac{10}{x^2} $ and $ g(x) = 0 $ represents the lower curve.

3. Substitute the equation for $ f(x) $ into the integral:
$$ V = \pi \int_{1}^{5} \left(\frac{10}{x^2}\right)^2 \, dx $$

4. Simplify the expression:
$$ V = \pi \int_{1}^{5} \frac{100}{x^4} \, dx $$

5. Integrate the expression:
$$ V = \pi \left[ -\frac{100}{3x^3} \right]_{1}^{5} $$

6. Evaluate the integral:
$$ V = \pi \left( -\frac{100}{3 \cdot 5^3} + \frac{100}{3 \cdot 1^3} \right) $$

$$ V = \pi \left( -\frac{100}{375} + \frac{100}{3} \right) $$

$$ V = \pi \left( -\frac{8}{15} + \frac{100}{3} \right) $$

$$ V = \pi \left( \frac{100}{3} - \frac{8}{15} \right) $$

$$ V = \pi \left( \frac{500}{15} - \frac{8}{15} \right) $$

$$ V = \pi \left( \frac{492}{15} \right) $$

$$ V \approx \frac{328 \pi}{5} $$

Therefore, the volume of the solid generated by revolving the region bounded by the given curves about the x-axis is approximately $ \frac{328 \pi}{5} $.