Find the area of the region bounded by the graphs of $y = 2x^2 + 3$ and $y = -2x + 7$. State the area of the region bounded by the 2 graphs if both graphs are translated 7 units in the negative $y$-direction. The region $R$ is bounded by $y = 2x^2 + 3$, $y = -2x + 7$, the $x$-axis and the $y$-axis. Find the exact volume of the solid generated when $R$ is rotated $2\pi$ about the $y$-axis.

1. Find the $x$-coordinates of the points of intersection between the two graphs.

$2x^2 + 3 = -2x + 7$
$2x^2 + 2x - 4 = 0$
$x^2 + x - 2 = 0$
$(x + 2)(x - 1) = 0$
$x = -2 \quad \text{or} \quad x = 1$

2. Set up the definite integral for the area between the two curves.

Since the line $y = -2x + 7$ is above the curve $y = 2x^2 + 3$ between $x = -2$ and $x = 1$, the area is:
$\int_{-2}^{1} [(-2x + 7) - (2x^2 + 3)] \, dx$
$= \int_{-2}^{1} (-2x^2 - 2x + 4) \, dx$

3. Evaluate the definite integral.

$= \left[ -\frac{2}{3}x^3 - x^2 + 4x \right]_{-2}^{1}$
$= \left( -\frac{2}{3}(1)^3 - (1)^2 + 4(1) \right) - \left( -\frac{2}{3}(-2)^3 - (-2)^2 + 4(-2) \right)$
$= \left( -\frac{2}{3} - 1 + 4 \right) - \left( \frac{16}{3} - 4 - 8 \right)$
$= \frac{7}{3} - \left( -\frac{20}{3} \right)$
$= \frac{27}{3}$
$= 9$

1. Find the $x$-coordinates of the points of intersection to determine the limits of integration for the area in (a).

$2x^2 + 3 = -2x + 7$
$2x^2 + 2x - 4 = 0$
$x^2 + x - 2 = 0$
$(x + 2)(x - 1) = 0$
$x = -2 \text{ or } x = 1$

2. Evaluate the definite integral to find the original area of the bounded region.

$\text{Area} = \int_{-2}^{1} [(-2x + 7) - (2x^2 + 3)] \, dx$
$= \int_{-2}^{1} (-2x^2 - 2x + 4) \, dx$
$= \left[ -\frac{2x^3}{3} - x^2 + 4x \right]_{-2}^{1}$
$= \left( -\frac{2}{3} - 1 + 4 \right) - \left( -\frac{16}{3} - 4 - 8 \right)$
$= \frac{7}{3} - \left(-\frac{20}{3}\right) = 9$

3. State the area of the translated region, recognising that translation preserves area.

A translation of both graphs by the same vector does not change the size or shape of the bounded region.
Therefore, the area remains unchanged from part (a).
$\text{Area} = 9$

1. Express both equations in terms of $y$ and determine the limits of integration.

From earlier parts, the line and curve intersect in the first quadrant at $x = 1$, giving $y = 5$.

For the line $y = -2x + 7$:
$2x = 7 - y \implies x = \frac{7-y}{2} \implies x^2 = \frac{(7-y)^2}{4}$
The line bounds the region down to the $x$-axis ($y=0$). Thus, the outer boundary of the solid spans from $y=0$ to $y=5$.

For the curve $y = 2x^2 + 3$:
$2x^2 = y - 3 \implies x^2 = \frac{y-3}{2}$
The curve's minimum is at $y=3$. Thus, the inner boundary (the "hole" in the solid) spans from $y=3$ to $y=5$.

2. Set up the definite integrals for the volume of revolution about the $y$-axis.

The solid's volume $V$ is generated by the line rotating about the $y$-axis, subtracting the volume generated by the curve.
$V = \pi \int_0^5 \left( x_{\text{line}} \right)^2 dy - \pi \int_3^5 \left( x_{\text{curve}} \right)^2 dy$
$V = \pi \int_0^5 \frac{(7-y)^2}{4} dy - \pi \int_3^5 \frac{y-3}{2} dy$

3. Evaluate the definite integral for the outer volume.

$V_{\text{outer}} = \frac{\pi}{4} \int_0^5 (7-y)^2 dy$
$V_{\text{outer}} = \frac{\pi}{4} \left[ \frac{-(7-y)^3}{3} \right]_0^5$
$V_{\text{outer}} = -\frac{\pi}{12} \left( (7-5)^3 - (7-0)^3 \right)$
$V_{\text{outer}} = -\frac{\pi}{12} (8 - 343) = \frac{335\pi}{12}$

4. Evaluate the definite integral for the inner volume and calculate the final exact volume.

$V_{\text{inner}} = \frac{\pi}{2} \int_3^5 (y-3) dy$
$V_{\text{inner}} = \frac{\pi}{2} \left[ \frac{(y-3)^2}{2} \right]_3^5$
$V_{\text{inner}} = \frac{\pi}{4} \left( (5-3)^2 - 0 \right) = \frac{\pi}{4} (4) = \pi$

$\text{Total Volume } V = V_{\text{outer}} - V_{\text{inner}} = \frac{335\pi}{12} - \frac{12\pi}{12} = \frac{323\pi}{12}$