### COMPUTING THE AREAS OF ENCLOSED REGIONS USING VERTICAL OR HORIZONTAL CROSS-SECTIONS

The following problems involve the use of integrals to compute the area of two-dimensional plane regions. Integration can use either vertical cross-sections or horizontal cross-sections. Often one method is easier than the other. Sometimes both methods work easily. We will begin with a brief review of describing regions in the plane using vertical and horizontal cross-sections.

DESCRIBING REGIONS IN THE PLANE USING VERTICAL OR HORIZONTAL CROSS-SECTIONS

EXAMPLE 1: Consider the region in the plane enclosed by the graphs of $y=2x$ and $y= x^2$. These graphs intersect when $x=0$ and $x=2$. Construct a vertical cross-section at $x$ for this region by FIRST picking a random value of $x$ between 0 and 2 and drawing a vertical line segment at $x$ starting from the graph of $y=x^2$ and ending on the graph of $y=2x$ (See the graph below.). This region can now be described using vertical cross-sections as follows : $$0 \le x \le 2 \ \ and \ \ x^2 \le y \le 2x$$ Note that the equations $y=x^2$ and $y=2x$ are equivalent to the equations $x= \sqrt{y}$ and $x= (1/2)y$, resp. Now construct a horizontal cross-section at $y$ for this region by FIRST picking a random value of $y$ between 0 and 4 and drawing a horizontal line segment at $y$ starting from the graph of $x= (1/2)y$ and ending on the graph of $x= \sqrt{y}$ (See the graph below.). This region can now be described using horizontal cross-sections as follows : $$0 \le y \le 4 \ \ and \ \ (1/2)y \le x \le \sqrt{y}$$

COMPUTING THE AREAS OF REGIONS IN THE PLANE USING INTEGRATION

To find the area of a region in the plane we simply integrate the height, $h(x)$, of a vertical cross-section at $x$ or the width, $w(y)$, of a horizontal cross-section at $y$. The generalization for finding areas of regions in the plane follows.

If a region $R$ in the plane can be described using vertical cross-sections as $$a \le x \le b \ \ and \ \ f(x) \le y \le g(x)$$ then the height of the vertical cross-section at $x$ is $h(x) = Top - Bottom = g(x) - f(x)$ (See figure below.)

and the area of $R$ is given by $$AREA = \int_a^b h(x) \, dx = \int_a^b ( g(x) - f(x) ) \, dx$$ If a region $R$ in the plane can be described using horizontal cross-sections as $$c \le y \le d \ \ and \ \ l(y) \le x \le r(y)$$ then the width of the horizontal cross-section at $y$ is $w(y) = Right - Left = r(y) - l(y)$ (See figure below.)

and the area of $R$ is given by $$AREA = \int_c^d w(y) \, dy = \int_c^d ( r(y) - l(y) ) \, dx$$ EXAMPLE 2: Compute the area of the region in the plane enclosed by the graphs of $y=2x$ and $y= x^2$. Using vertical cross-sections we have that $$0 \le x \le 2 \ \ and \ \ x^2 \le y \le 2x$$ so that the area of the region is $$AREA = \int_0^2 h(x) \, dx = \int_0^2 ( Top - Bottom ) \, dx$$ $$= \int_0^2 ( 2x-x^2 ) \, dx$$ $$= (x^2-(1/3)x^3) \Big\vert_{0}^{2}$$ $$= 4 - 8/3$$ $$= 4/3$$ Using horizontal cross-sections we have that $$0 \le y \le 4 \ \ and \ \ (1/2)y \le x \le \sqrt{y}$$ so that the area of the region is $$AREA = \int_0^4 w(y) \, dy = \int_0^4 ( Right - Left ) \, dy$$ $$= \int_0^4 ( \sqrt{y} - (1/2)y ) \, dy$$ $$= ( (2/3)x^{3/2}- (1/4)y^2 ) \Big\vert_{0}^{4}$$ $$= (2/3)4^{3/2}- (1/4)4^2$$ $$= (2/3)(8)- (1/4)(16)$$ $$= 16/3- 4$$ $$= 4/3$$

The following problems use integration to find areas of regions in the plane. Most are average. A few are somewhat challenging. It's arguable that the most important component in getting the correct solutions to the following problems is a carefully and completely detailed graph of the enclosed region. This includes proper labeling of the axes, graphs, intercepts, and cross-sections.

Compute the area of the region enclosed by the graphs of the given equations. Use vertical cross-sections on Problems 1-16.

• PROBLEM 1 : $y=x, y=2x,$ and $x=4$

• PROBLEM 2 : $y=x^2$ and $y=x+2$

• PROBLEM 3 : $y=e^x, y=e^{-x},$ and $x= \ln 3$

• PROBLEM 4 : $y=x^2$ and $y=-x^2+4x+6$

• PROBLEM 5 : $y=x^3+x^2$ and $y=3x^2+3x$

• PROBLEM 6 : $y= \ln x, y=1,$ and $x=e^2$

• PROBLEM 7 : $y= \cos x, y= \sin x,$ and $x=0$

• PROBLEM 8 : $y=8/x, y=2x,$ and $y=2$

• PROBLEM 9 : $y=\ln x$ and $y= (\ln x)^2$

• PROBLEM 10 : $y=\tan^2 x, y=0,$ and $x=1$

• PROBLEM 11 : $y=x, y=2x,$ and $y=6-x$

• PROBLEM 12 : $y=\sin \sqrt{x}$ and $y=0$ on the interval $[0, \pi^2]$

• PROBLEM 13 : $y=x^3$ and $y=4x$

• PROBLEM 14 : $y=\displaystyle \frac{1}{4}x^4-x^2$ and $y=x(x-2)(x+2)$

• PROBLEM 15 : $y=\sin x$ and $\sin 2x$ on the interval $[0, \pi]$

• PROBLEM 16 : $y=2x$ and $y= \displaystyle{ 1 \over 2 }x-4$ , $y=0$, and $y=2$.

Compute the area of the region enclosed by the graphs of the given equations. Use horizontal cross-sections on Problems 17-22.

• PROBLEM 17 : $x=y^2$ and $x=4$

• PROBLEM 18 : $x=y+3$ and $x=y^2-y$

• PROBLEM 19 : $x=y^3$ and $x=y^2+2y$

• PROBLEM 20 : $y= \displaystyle\frac{2}{x}, y= \frac{1}{2}x,$ and $y=2$

• PROBLEM 21 : $y= \tan x, y=x,$ and BELOW $y=3$

• PROBLEM 22 : $y=2x$ and $y= \displaystyle{ 1 \over 2 }x-4$ , $y=0$, and $y=2$

In Problems 23-28 compute the area of the region enclosed by the graphs of the given equations. Use a.) vertical cross-sections and b.) horizontal cross-sections. SET UP BUT DO NOT EVALUATE THE INTEGRALS.
• PROBLEM 23 : $y= 3x, y=x,$ and $x=2$

• PROBLEM 24 : $y= 2x, y=\displaystyle \frac{1}{2}x,$ and $y=4$

• PROBLEM 25 : $x=y^2$ and $x=y+2$

• PROBLEM 26 : $y=e^x, y=e^{-x},$ and $x=2$

• PROBLEM 27 : $y= \ln x, y=1-x,$ and $y=2$

• PROBLEM 28 : $y=x^3, y=x+6, y=2x-6,$ and $y=0$

• PROBLEM 29 : $y=x^2-2x+2$ and $y=2x+2$