Question 10
The region $R$, shown shaded in Figure 2, is bounded by the curve with equation $y = x^2 + 1$ and the curve with equation $x^2 + y^2 = 11$.
The two curves intersect at the point $A$ and at the point $B$.
(a) Find the $x$ coordinate of the point $A$ and the $x$ coordinate of the point $B$.
The region $R$ is rotated through $360^\circ$ about the $x$-axis.
(b) Use algebraic integration to find the volume, to 2 decimal places, of the solid generated.
Solution
Step 1: Finding the $x$ coordinates of the points of intersection
We are given two curves:
$$ y = x^2 + 1 \quad \text{and} \quad x^2 + y^2 = 11. $$
To find the $x$ coordinates of the points of intersection, substitute the expression for $y$ from the first equation into the second equation:
$$ x^2 + (x^2 + 1)^2 = 11. $$
Expanding the square:
$$ x^2 + (x^4 + 2x^2 + 1) = 11. $$
Simplifying:
$$ x^4 + 3x^2 + 1 = 11. $$
Subtracting 11 from both sides:
$$ x^4 + 3x^2 - 10 = 0. $$
Let $z = x^2$. Then the equation becomes:
$$ z^2 + 3z - 10 = 0. $$
Solve this quadratic equation using the quadratic formula:
$$ z = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2}. $$
Thus:
$$ z = \frac{-3 \pm 7}{2}. $$
The two possible solutions for $z$ are:
$$ z = \frac{-3 + 7}{2} = 2 \quad \text{or} \quad z = \frac{-3 - 7}{2} = -5. $$
Since $z = x^2$, and $x^2$ cannot be negative, we take $z = 2$. Therefore, $x^2 = 2$, giving:
$$ x = \pm \sqrt{2}. $$
Thus, the $x$ coordinates of the points of intersection are $x = -\sqrt{2}$ and $x = \sqrt{2}$.
Step 2: Finding the volume of the solid generated
The region $R$ is rotated about the $x$-axis. The volume of the solid can be calculated using the method of cylindrical shells or disks. Here, we use the disk method.
The volume of the solid generated by rotating the region between $y = x^2 + 1$ and the circle $x^2 + y^2 = 11$ about the $x$-axis is given by:
$$ V = \pi \int_{-\sqrt{2}}^{\sqrt{2}} \left[ (11 - x^2) - (x^2 + 1)^2 \right] \, dx. $$
First, simplify the integrand:
$$ (x^2 + 1)^2 = x^4 + 2x^2 + 1, $$
and
$$ 11 - x^2 = 11 - x^2. $$
Thus, the integrand becomes:
$$ -(x^4 + 2x^2 + 1) + (11 - x^2) = -(x^4 + 3x^2 - 10). $$
Therefore, the volume integral is:
$$ V = \pi \int_{-\sqrt{2}}^{\sqrt{2}} -(x^4 + 3x^2 - 10) \, dx. $$
Since the integrand is an even function, we can simplify the integral by doubling the integral from $0$ to $\sqrt{2}$:
$$ V = 2\pi \int_0^{\sqrt{2}} -(x^4 + 3x^2 - 10) \, dx. $$
Now, integrate term by term:
$$ \int x^4 \, dx = \frac{x^5}{5}, \quad \int 3x^2 \, dx = x^3, \quad \int -10 \, dx = -10x. $$
Thus, the volume integral becomes:
$$ V = -2\pi \left[ \frac{x^5}{5} + x^3 - 10x \right]_0^{\sqrt{2}}. $$
Evaluating at the upper limit $x = \sqrt{2}$:
$$ -\frac{(\sqrt{2})^5}{5} - (\sqrt{2})^3 + 10(\sqrt{2}) = -\frac{4\sqrt{2}}{5} - 2\sqrt{2} + 10\sqrt{2}. $$
Simplifying:
$$ -\frac{4\sqrt{2}}{5} - 2\sqrt{2} + 10\sqrt{2} = \left( -\frac{4}{5} - 2 + 10 \right)\sqrt{2} = \left( -\frac{4}{5} + 8 \right)\sqrt{2} = \frac{36}{5}\sqrt{2}. $$
Thus, the volume is:
$$ V = 2\pi \left( \frac{36}{5}\sqrt{2} \right) = \frac{72}{5}\pi\sqrt{2} = 63.977. $$
The final answer for the volume is approximately:
$$ V \approx 63.98 \, \text{cubic units}. $$
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