The diagram below shows a region $R$ bounded by the curve $(y+5)^{2}=x-3$ and the line $y=x-10$. Find the volume of solid formed when $R$ is rotated through four right angles about the x-axis.

1. Find the points of intersection between the curve and the line.

Substitute the line equation $y = x - 10 \implies x = y + 10$ into the curve's equation:
$(y+5)^2 = (y+10) - 3$
$y^2 + 10y + 25 = y + 7$
$y^2 + 9y + 18 = 0$
$(y+3)(y+6) = 0$
$y = -3 \quad \text{or} \quad y = -6$
When $y = -3$, $x = 7$. When $y = -6$, $x = 4$.
Thus, the intersections are $(4, -6)$ and $(7, -3)$.

2. Express the curve with $y$ as the subject and identify the upper and lower branches to determine the boundaries of the region.

Making $y$ the subject of the curve's equation:
$(y+5)^2 = x - 3 \implies y = -5 \pm \sqrt{x-3}$
The parabola opens to the right with its vertex at $(3, -5)$.

The inner boundary of the region $R$ (closer to the $x$-axis, hence smaller radius of revolution) is the upper branch:
$y_{1} = -5 + \sqrt{x-3} \quad \text{for } x \in [3, 7]$

The outer boundary (further from the $x$-axis, hence larger radius) is split into two sections:

1. The lower branch of the curve: $y_{2} = -5 - \sqrt{x-3}$ for $x \in [3, 4]$
2. The line: $y_{3} = x - 10$ for $x \in [4, 7]$

3. Set up the definite integrals for the volume of the solid generated when the region $R$ is rotated $360^\circ$ about the $x$-axis.

The volume $V$ is found using the washer/disc method, requiring us to split the region at $x=4$ based on the outer boundaries:
$V = \pi \int_{3}^{4} \left( y_{2}^2 - y_{1}^2 \right) \mathrm{d}x + \pi \int_{4}^{7} \left( y_{3}^2 - y_{1}^2 \right) \mathrm{d}x$

Substituting the functions:
$V_1 = \pi \int_{3}^{4} \left[ (-5 - \sqrt{x-3})^2 - (-5 + \sqrt{x-3})^2 \right] \mathrm{d}x$
$V_2 = \pi \int_{4}^{7} \left[ (x-10)^2 - (-5 + \sqrt{x-3})^2 \right] \mathrm{d}x$

4. Evaluate the integrals (analytically or via a graphing calculator) to find the exact volume.

Simplifying the integrands:
$V_1 = \pi \int_{3}^{4} 20\sqrt{x-3} \, \mathrm{d}x$
$V_2 = \pi \int_{4}^{7} \left( x^2 - 21x + 78 + 10\sqrt{x-3} \right) \mathrm{d}x$

Evaluating the definite integrals:
$V_1 = \pi \left[ \frac{40}{3}(x-3)^{\frac{3}{2}} \right]_3^4 = \frac{40}{3}\pi$
$V_2 = \pi \left[ \frac{x^3}{3} - \frac{21x^2}{2} + 78x + \frac{20}{3}(x-3)^{\frac{3}{2}} \right]_4^7 = \frac{163}{6}\pi$

Total volume:
$V = V_1 + V_2 = \frac{40}{3}\pi + \frac{163}{6}\pi = \frac{243}{6}\pi = \frac{81\pi}{2}$

1. Set up the first application of integration by parts.

Let $I = \int e^{-2x}\cos x\,\mathrm{d}x$.
Using the integration by parts formula $\int u\,\mathrm{d}v = uv - \int v\,\mathrm{d}u$:
Let $u = e^{-2x}$ and $\mathrm{d}v = \cos x\,\mathrm{d}x$.
Then $\mathrm{d}u = -2e^{-2x}\,\mathrm{d}x$ and $v = \sin x$.

2. Apply the integration by parts formula to obtain a new integral.

$I = e^{-2x}\sin x - \int (\sin x)(-2e^{-2x})\,\mathrm{d}x$
$I = e^{-2x}\sin x + 2\int e^{-2x}\sin x\,\mathrm{d}x$

3. Apply integration by parts a second time to the new integral.

For $\int e^{-2x}\sin x\,\mathrm{d}x$, let $u = e^{-2x}$ and $\mathrm{d}v = \sin x\,\mathrm{d}x$.
Then $\mathrm{d}u = -2e^{-2x}\,\mathrm{d}x$ and $v = -\cos x$.

$\int e^{-2x}\sin x\,\mathrm{d}x = -e^{-2x}\cos x - \int (-\cos x)(-2e^{-2x})\,\mathrm{d}x$
$\int e^{-2x}\sin x\,\mathrm{d}x = -e^{-2x}\cos x - 2\int e^{-2x}\cos x\,\mathrm{d}x$

4. Substitute back to form an equation in terms of $I$ and solve.

Substitute this result back into the earlier equation for $I$:
$I = e^{-2x}\sin x + 2(-e^{-2x}\cos x - 2I)$
$I = e^{-2x}\sin x - 2e^{-2x}\cos x - 4I$
$5I = -2e^{-2x}\cos x + e^{-2x}\sin x$
$I = -\dfrac{2}{5}e^{-2x}\cos x + \dfrac{1}{5}e^{-2x}\sin x + C$
(where $C$ is an arbitrary constant of integration.)

1. Find the value of parameter $t$ and coordinates at the $y$-axis.

At the $y$-axis, $x = 0$.
$\sin t = 0 \implies t = 0$
When $t = 0$, $y = e^{-2(0)} = 1$.
The point of intersection with the $y$-axis is $(0, 1)$.

2. Find the gradient of the curve at $t=0$.

Using parametric differentiation:
$\frac{\mathrm{d}x}{\mathrm{d}t} = \cos t$
$\frac{\mathrm{d}y}{\mathrm{d}t} = -2e^{-2t}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{-2e^{-2t}}{\cos t}$
At $t = 0$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-2e^0}{\cos 0} = -2$

3. Determine the equation of the tangent at the $y$-axis.

Using the point $(0, 1)$ and gradient $m = -2$:
$y - 1 = -2(x - 0)$
$y = -2x + 1$

4. Set up the area integral for the region.

The region is bounded by the curve, the tangent, and the line $x = 1$. The integration bounds for $x$ are from $0$ to $1$.
$\text{Area} = \int_{0}^{1} (y_{\text{curve}} - y_{\text{tangent}}) \,\mathrm{d}x$
$\text{Area} = \int_{0}^{1} y_{\text{curve}} \,\mathrm{d}x - \int_{0}^{1} (-2x + 1) \,\mathrm{d}x$

5. Evaluate the area using parametric substitution and the result from (b)(i).

For the curve integral, substitute $x = \sin t \implies \mathrm{d}x = \cos t \,\mathrm{d}t$. As $x$ goes from $0$ to $1$, $t$ goes from $0$ to $\frac{\pi}{2}$.
$\int_{0}^{1} y_{\text{curve}} \,\mathrm{d}x = \int_{0}^{\frac{\pi}{2}} e^{-2t} \cos t \,\mathrm{d}t$
Using the integration result from (b)(i):
$\int_{0}^{\frac{\pi}{2}} e^{-2t} \cos t \,\mathrm{d}t = \left[ -\frac{2}{5}e^{-2t}\cos t + \frac{1}{5}e^{-2t}\sin t \right]_0^{\frac{\pi}{2}}$
$= \left( 0 + \frac{1}{5}e^{-\pi} \right) - \left( -\frac{2}{5} + 0 \right) = \frac{1}{5}e^{-\pi} + \frac{2}{5}$
Now evaluate the tangent line's definite integral:
$\int_{0}^{1} (-2x + 1) \,\mathrm{d}x = \left[ -x^2 + x \right]_0^1 = (-1 + 1) - 0 = 0$
Therefore, the exact area is:
$\left( \frac{1}{5}e^{-\pi} + \frac{2}{5} \right) - 0 = \frac{1}{5}e^{-\pi} + \frac{2}{5}$