To find the volume of the solid obtained by rotating the region enclosed by the curves $x = y^4$ and $x = \sqrt[4]{y}$ about the $y$-axis, we can use the method of cylindrical shells or the method of washers/disks. Here, I'll use the method of washers.

Step 1: Find the points of intersection

The first step is to determine where the two curves intersect.

$y^4 = \sqrt[4]{y}$

Taking the fourth power of both sides to eliminate the roots:

$y^{16} = y$

This simplifies to:

$y^{16} - y = 0$

Factoring out $y$:

$y(y^{15} - 1) = 0$

This gives:

$y = 0 \quad \text{or} \quad y^{15} = 1$

For $y^{15} = 1$, we find:

$y = 1$

So, the curves intersect at $y = 0$ and $y = 1$.

Step 2: Set up the volume integral

The volume $V$ of the solid of revolution about the $y$-axis using the method of washers is given by:

$V = \pi \int_{a}^{b} \left[ R_{\text{outer}}^2(y) - R_{\text{inner}}^2(y) \right] \, dy$

Here:

• The outer radius $R_{\text{outer}}(y) = \sqrt[4]{y}$
• The inner radius $R_{\text{inner}}(y) = y^4$
• The bounds are from $y = 0$ to $y = 1$.

So the volume integral becomes:

[ V = \pi \int_{0}^{1} \left[ \left(\sqrt[4]{y}\right)^2 - \left(y^4\right)^2 \right] , dy ]

Step 3: Simplify and integrate

Simplify the integrand:

$V = \pi \int_{0}^{1} \left[ y^{1/2} - y^8 \right] \, dy$

Now, integrate term by term:

$V = \pi \left[ \int_{0}^{1} y^{1/2} \, dy - \int_{0}^{1} y^8 \, dy \right]$

Integrate each term:

1. $\int y^{1/2} \, dy = \frac{y^{3/2}}{\frac{3}{2}} = \frac{2}{3} y^{3/2}$
2. $\int y^8 \, dy = \frac{y^9}{9}$

Evaluating these at the bounds $y = 0$ to $y = 1$:

$V = \pi \left[ \left( \frac{2}{3} y^{3/2} \right) \Big|_{0}^{1} - \left( \frac{y^9}{9} \right) \Big|_{0}^{1} \right]$

$V = \pi \left[ \frac{2}{3} (1) - \frac{1}{9} (1) \right]$

$V = \pi \left[ \frac{2}{3} - \frac{1}{9} \right]$

Find a common denominator:

$V = \pi \left[ \frac{6}{9} - \frac{1}{9} \right]$

$V = \pi \cdot \frac{5}{9}$

$V = \frac{5\pi}{9}$

Final Answer:

The volume of the solid obtained by rotating the region enclosed by the graphs $x = y^4$ and $x = \sqrt[4]{y}$ in the first quadrant about the $y$-axis is $\frac{5\pi}{9}$.

Would you like any more details on the solution?

Here are 5 related questions:

1. How does the method of cylindrical shells work for finding volumes of solids of revolution?
2. What is the difference between using the disk method and the washer method in volume calculations?
3. How do you find the points of intersection for more complex curves?
4. What are some real-life applications of solids of revolution?
5. Can the same method be used to find the volume if the solid is rotated about the x-axis instead?

Tip: Always double-check the integration bounds and the radii used in your integral, as these are common sources of error in volume problems.