Question 2

Let $\mathcal{R}$ be the region in the plane bounded by the curves:

$$y = 16 - x^2\hspace{4pt} \textrm{and}\hspace{4pt}y = x^2 - 4x.$$

Sketch the region $\mathcal{R}$ and find the area of the region $\mathcal{R}$.

The two curves meet where both equations hold at once, so where:

$$16 - x^2 = x^2 - 4x,$$

$$0 = 2x^2- 4x - 16,$$

$$0 = x^2 - 2x - 8,$$

$$0 = (x - 4)(x + 2).$$

So they meet at $x = 4$, where $y = 0$ for both curves and at $x = - 2$, where $y = 12$ for both curves.
The region $\mathcal{R}$ is then the region below the concave down parabola $y_+ = 16 -x^2$ and above the concave up parabola $y_- = x^2 - 4x$, in the $x$-interval $[-2, 4]$.

Using vertical slivers, of width $dx$ and height $y_+ - y_-$, the area element $d\mathcal{A}$ is:

$$d\mathcal{A} = (y_+ - y_-) \,\,dx = (16 - x^2 - (x^2 - 4x))\,\,dx = (16 - 2x^2 + 4x )\,\,dx.$$

So the required total area $\mathcal{A}$ is:

$$\mathcal{A} = \int_{x = -2}^{x = 4} d\mathcal{A} = \int_{-2}^4 (16 - 2x^2 + 4x )\,\,dx$$

$$= [16x - \frac{2x^3}{3} + 2x^2]_{-2}^4$$

$$= (16(4) - \frac{2(4^3)}{3} + 2(4^2)] - (16(-2) - \frac{2((-2)^3)}{3} + 2((-2)^2))$$

$$= 64 - \frac{128}{3} + 32 - (-32 + \frac{16}{3} + 8)$$

$$= 96 - \frac{128}{3} + 32 - \frac{16}{3} - 8$$

$$= 120 - \frac{144}{3}$$

$$= 120 - 48 = 72.$$