Let $$ R $$ be the region bounded by the curves $$ y=x $$ and $$ y=x^{\frac{1}{3}} $$. Let $$ S $$ be the solid generated when $$ R $$ is revolved about the $$ x $$-axis in the first quadrant. Find the volume of $$ S $$ by both the disc/washer and shell methods. Check that your results agree.

Question

UpStudy AI · Accepted Answer

To find the volume of the solid $$ S $$ generated by revolving the region $$ R $$ bounded by the curves $$ y = x $$ and $$ y = x^{\frac{1}{3}} $$ about the $$ x $$-axis, we can use both the **washer (disc) method** and the **shell method**. Let's proceed step-by-step.

### 1. Identifying the Region $$ R $$

First, determine the points of intersection between $$ y = x $$ and $$ y = x^{\frac{1}{3}} $$:

$$
x = x^{\frac{1}{3}} \implies x^3 = x \implies x(x^2 - 1) = 0 \implies x = 0 \text{ or } x = \pm 1
$$

Since we're considering the first quadrant ($$ x \geq 0 $$), the intersection points are $$ x = 0 $$ and $$ x = 1 $$. For $$ 0 \leq x \leq 1 $$:

$$
y = x^{\frac{1}{3}} \geq y = x
$$

Thus, the region $$ R $$ is bounded by $$ x = 0 $$, $$ x = 1 $$, $$ y = x $$, and $$ y = x^{\frac{1}{3}} $$.

### 2. Washer (Disc) Method

When revolving around the $$ x $$-axis, the volume $$ V $$ can be computed using:

$$
V = \pi \int_{a}^{b} \left[ (y_{\text{upper}})^2 - (y_{\text{lower}})^2 \right] dx
$$

Here:

- Upper curve: $$ y = x^{\frac{1}{3}} $$
- Lower curve: $$ y = x $$
- Limits of integration: $$ a = 0 $$, $$ b = 1 $$

Thus,

$$
V = \pi \int_{0}^{1} \left[ \left( x^{\frac{1}{3}} \right)^2 - x^2 \right] dx = \pi \int_{0}^{1} \left( x^{\frac{2}{3}} - x^2 \right) dx
$$

Calculating the integral:

$$
\int x^{\frac{2}{3}} dx = \frac{3}{5} x^{\frac{5}{3}}, \quad \int x^2 dx = \frac{1}{3} x^3
$$

Plugging in the limits:

$$
V = \pi \left[ \frac{3}{5}x^{\frac{5}{3}} - \frac{1}{3}x^3 \right]_0^1 = \pi \left( \frac{3}{5} - \frac{1}{3} \right) = \pi \left( \frac{9}{15} - \frac{5}{15} \right) = \pi \left( \frac{4}{15} \right) = \frac{4\pi}{15}
$$

### 3. Shell Method

For the shell method, it's convenient to integrate with respect to $$ y $$. Express $$ x $$ in terms of $$ y $$:

$$
y = x \implies x = y, \quad y = x^{\frac{1}{3}} \implies x = y^3
$$

The height of each cylindrical shell is $$ x_{\text{right}} - x_{\text{left}} = y - y^3 $$, and the radius is $$ y $$. The volume $$ V $$ is given by:

$$
V = 2\pi \int_{c}^{d} (\text{radius})(\text{height}) dy = 2\pi \int_{0}^{1} y (y - y^3) dy
$$

Simplifying the integral:

$$
V = 2\pi \int_{0}^{1} (y^2 - y^4) dy = 2\pi \left[ \frac{1}{3}y^3 - \frac{1}{5}y^5 \right]_0^1 = 2\pi \left( \frac{1}{3} - \frac{1}{5} \right) = 2\pi \left( \frac{2}{15} \right) = \frac{4\pi}{15}
$$

### 4. Conclusion

Both methods yield the same result:

$$
\boxed{\dfrac{4\,\pi}{15}}
$$
The volume of the solid $$ S $$ is $$ \frac{4\pi}{15} $$ cubic units.

Let be the region bounded by the curves and . Let be the solid generated
when is revolved about the -axis in the first quadrant. Find the volume of by both the
disc/washer and shell methods. Check that your results agree.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Mind Expander

Volume Using the Disc/Washer Method

Volume Using the Shell Method

Conclusion

preguntas relacionadas

Latest Calculus Questions

Let be the region bounded by the curves and . Let be the solid generated when is revolved about the -axis in the first quadrant. Find the volume of by both the disc/washer and shell methods. Check that your results agree.