Solutions to Problem Set 11

Math 211-03

9-26-2017

[Area]

1. Find the area of the region bounded by $y = 2 x^2 - 8 x + 3$ and $y
   = x^2 - 2 x - 2$

$$\hbox{\epsfysize=1.5in \epsffile{area10.eps}}$$

Find the intersections:

$$\eqalign{ 2 x^2 - 8 x + 3 & = x^2 - 2 x - 2 \cr x^2 - 6 x + 5 & = 0 \cr (x - 1)(x - 5) & = 0 \cr}$$

The curves intersect at $x = 1$ and $x = 5$ . Between 1 and 5, the top curve is $y = x^2 - 2 x -
   2$ and the bottom curve is $y = 2 x^2 - 8 x + 3$ . The area is

$$\int_1^5 [(x^2 - 2 x - 2) - (2 x^2 - 8 x + 3)]\,dx = \int_1^5 (-x^2 + 6 x - 5)\,dx =$$

$$\left[-\dfrac{1}{3} x^3 + 3 x^2 - 5 x\right]_1^5 = \dfrac{32}{3} = 10.66666 \ldots.\quad\halmos$$


2. Find the area of the region between $y = 8 - x$ and $y = 3 x$ from $x = 0$ to $x = 3$ .

$$\hbox{\epsfysize=1.5in \epsffile{area9.eps}}$$

The lines cross in the middle of the interval. To find the intersection point, solve simultaneously:

$$\eqalign{ 8 - x & = 3 x \cr 8 & = 4 x \cr 2 & = x \cr}$$

The lines cross at $x = 2$ .

From $x = 0$ to $x = 2$ , the line $y = 8 - x$ is the top curve and the line $y = 3 x $ is the bottom curve. From $x = 2$ to $x = 3$ , the line $y = 3 x$ is the top curve and the line $y = 8 - x$ is the bottom curve. The area is

$$\int_0^2 (8 - x - 3 x)\,dx + \int_2^3 [3 x - (8 - x)]\,dx = \int_0^2 (8 - 4 x)\,dx + \int_2^3 (4 x - 8)\,dx =$$

$$\left[8 x - 2 x^2\right]_0^2 + \left[2 x^2 - 8 x\right]_2^3 = 10.\quad\halmos$$


3. Find the area of the region between the curves $y = x^2 - 4$ and $y = -3 x$ from $x = 0$ to $x = 3$ .

$$\hbox{\epsfysize=1.75in \epsffile{area4.eps}}$$

The curves cross in the middle of the interval. To find the intersection point, solve simultaneously:

$$\eqalign{ x^2 - 4 & = -3 x \cr x^2 + 3 x - 4 & = 0 \cr (x + 4)(x - 1) & = 0 \cr}$$

This gives $x = -4$ and $x
   = 1$ . The intersection point between 0 and 3 is at $x = 1$ .

I'll use vertical rectangles to compute the area.

For the left-hand piece, the top curve is $y = -3 x$ and the bottom curve is $y = x^2 - 4$ .

For the right-hand piece, the top curve is $y = x^2 - 4$ and the bottom curve is $y = -3 x$ .

The total area is

$$\int_0^1 \left(-3 x - (x^2 - 4)\right)\,dx + \int_1^3 \left((x^2 - 4) - (-3 x)\right)\,dx = \int_0^1 \left(4 - 3 x - x^2\right)\,dx + \int_1^3 \left(x^2 + 3 x - 4\right)\,dx =$$

$$\left[4 x - \dfrac{3}{2} x^2 - \dfrac{1}{3} x^3\right]_0^1 + \left[\dfrac{1}{3} x^3 + \dfrac{3}{2} x^2 - 4 x\right]_1^3 = \dfrac{89}{6} \approx 14.83333.\quad\halmos$$


4. Find the area of the shaded region in the picture below.

$$\hbox{\epsfysize=2in \epsffile{area5.eps}}$$

It's possible to compute the area using vertical rectangles, but you'll need two integrals to do it. I'll use horizontal rectangles instead.

$$\hbox{\epsfysize=2in \epsffile{area5b.eps}}$$

The thickness of a horizontal rectangle is $dy$ , so I want to express my integrand in terms of y.

The right-hand end of a horizontal rectangle lies on $y = 4 - x^2$ , which is $x =
   \sqrt{4 - y}$ . The left-hand end lies on $y = 3 x$ , which is $x = \dfrac{y}{3}$ .

Set the curves equal to find the intersection point:

$$4 - x^2 = 3 x, \quad x^2 + 3 x - 4 = 0, \quad (x + 4)(x - 1) = 0, \quad x = -4 \quad\hbox{or}\quad x = 1.$$

When $x = 1$ , $y = 3 x = 3$ . So the limits on y are $y = 0$ (the x-axis) and $y = 3$ (the intersection point). The area is

$$\int_0^3 \left(\sqrt{4 - y} - \dfrac{y}{3}\right)\,dy = \left[-\dfrac{2}{3} (4 - y)^{3/2} - \dfrac{y^2}{6}\right]_0^3 = \dfrac{19}{6} \approx 3.16667.\quad\halmos$$


5. Find the area of the region bounded by $x = y^2 - 4 y + 6$ and $x
   = 2 + 2 y - y^2 $ .

$$\hbox{\epsfysize=1.75in \epsffile{area6.eps}}$$

Use horizontal rectangles. The right-hand end of a horizontal rectangle lies on $x = 2 + 2 y -
   y^2$ and the left-hand end lies on $x = y^2 - 4 y + 6$ .

Find the intersection points:

$$2 + 2 y - x^2 = y^2 - 4 y + 6, \quad 2 y^2 - 6 y + 4 = 0, \quad y^2 - 3 y + 2 = 0, \quad (y - 1)(y - 2) = 0, \quad y = 1 \quad\hbox{or}\quad y = 2.$$

The area is

$$\int_1^2 \left((2 + 2 y - y^2) - (y^2 - 4 y + 6)\right)\,dy = \int_1^2 \left(6 y - 4 - 2 y^2\right)\,dy = \left[3 y^2 - 4 y - \dfrac{2}{3} y^3\right]_1^2 = \dfrac{1}{3}.\quad\halmos$$


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