To find the volume of the solid obtained by rotating the region bounded by the curves $x = y^2$ and $x = 1 - y^2$ about the line $x = 3$, we can use the washer method.

Step 1: Sketch the region and understand the bounds

1. The curves are:

   • $x = y^2$: A parabola opening to the right.
   • $x = 1 - y^2$: A parabola opening to the left.

2. Find the points of intersection of the two curves: $y^2 = 1 - y^2$ $2y^2 = 1 \implies y^2 = \frac{1}{2} \implies y = \pm\frac{1}{\sqrt{2}}$

Step 2: Set up the washer method

To find the volume, we calculate the area of a representative washer and integrate with respect to $y$. The radii of the washers are determined by the distances from the curves to the line $x = 3$.

• Outer radius $R(y)$: Distance from $x = 3$ to the leftmost curve $x = y^2$.

  $R(y) = 3 - y^2$

• Inner radius $r(y)$: Distance from $x = 3$ to the rightmost curve $x = 1 - y^2$.

  $r(y) = 3 - (1 - y^2) = 2 + y^2$

Step 3: Set up the integral

The volume of the solid is given by:

$V = \pi \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left[(R(y))^2 - (r(y))^2\right] \, dy$

Substituting the expressions for the radii:

$V = \pi \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left[(3 - y^2)^2 - (2 + y^2)^2\right] \, dy$

Step 4: Simplify the integrand

Let's expand and simplify the expressions inside the integral:

1. Expand $(3 - y^2)^2$: $(3 - y^2)^2 = 9 - 6y^2 + y^4$

2. Expand $(2 + y^2)^2$: $(2 + y^2)^2 = 4 + 4y^2 + y^4$

Subtract the inner expansion from the outer:

$(3 - y^2)^2 - (2 + y^2)^2 = (9 - 6y^2 + y^4) - (4 + 4y^2 + y^4) = 5 - 10y^2$

Step 5: Evaluate the integral

$V = \pi \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (5 - 10y^2) \, dy$

Since the integrand is even, we can simplify the integral:

$V = 2\pi \int_{0}^{\frac{1}{\sqrt{2}}} (5 - 10y^2) \, dy$

Now evaluate this:

$V = 2\pi \left[ 5y - \frac{10y^3}{3} \right]_0^{\frac{1}{\sqrt{2}}}$

Substituting the limits:

$V = 2\pi \left[ 5\left(\frac{1}{\sqrt{2}}\right) - \frac{10\left(\frac{1}{\sqrt{2}}\right)^3}{3} \right]$

Simplify:

[ V = 2\pi \left[ \frac{5}{\sqrt{2}} - \frac{10}{3\sqrt{8}} \right] = 2\pi \left[ \frac{5}{\sqrt{2}} - \frac{10}{3 \cdot 2\sqrt{2}} \right] ]

[ V = 2\pi \left[ \frac{5}{\sqrt{2}} - \frac{10}{6\sqrt{2}} \right] = 2\pi \left[ \frac{30}{6\sqrt{2}} - \frac{10}{6\sqrt{2}} \right] = 2\pi \cdot \frac{20}{6\sqrt{2}} ]

$V = \frac{20\pi}{3\sqrt{2}} = \frac{20\pi\sqrt{2}}{6} = \frac{10\pi\sqrt{2}}{3}$

Final Volume

$V = \frac{10\pi\sqrt{2}}{3}$

Let me know if you have any questions or need further details!

Related Questions

1. How do you find the volume using the shell method?
2. What happens if the rotation axis changes to another line?
3. How do you calculate the volume if the region is rotated around the y-axis?
4. Can this approach be applied to other shapes and curves?
5. What are the limitations of the washer method?

Tip: Always draw the region and the axis of rotation to visualize the solid formed; this helps in setting up the correct radii for the washer method.