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Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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There are several topics in calculus where you need to be able to describe an area in the plane. This skill is necessary for calculating volume using single integrals as well area and volumes for double and triple integrals.

To describe an area in the xy-plane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot ( used to plot these graphs; we used gimp to add labels and other graphics ). However, graphing by hand is usually the best and quickest way.

We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by \(f(x)\) (red line), \(g(x)\) (blue line) and \(x=a\) (black line).
[Remember that an equation like \(x=a\) can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]

Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).

Vertically

Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.

Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing \(g(x)\), no matter where we draw it. Similarly, the arrow always exits the area by crossing \(f(x)\), no matter where we draw it. Do you see that?

But wait, how far to the right does it go? We are not given that information. What we need to do is find the x-value where the functions \(f(x)\) and \(g(x)\) intersect. You should be able to do that. We will call that point \((b,f(b))\). Also, we will call the left boundary \(x=a\). So now we have everything we need to describe this area. We give the final results below.

Vertical Arrow

\( g(x) \leq y \leq f(x) \)

arrow leaves through \(f(x)\) and enters through \(g(x)\)

\( a \leq x \leq b \)

arrow sweeps from left (\(x=a\)) to right (\(x=b\))

Horizontally

We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of \(f(x)\) and \(g(x)\) in terms of \(y\). ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations \(f(x) \to F(y)\) and \(g(x) \to G(y)\).

Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line \(x=a\). However, there is something strange going on at the point \((b,f(b))\). Notice that when the arrow is below \(f(b)\), the arrow exits through \(g(x)\) but when the arrow is above \(f(b)\), the arrow exits through \(f(x)\). This is a problem. To overcome this, we need to break the area into two parts at \(f(b)\).

Lower Section - - This section is described by the arrow leaving through \(g(x)\). So the arrow sweeps from \(g(a)\) to \(g(b)\).
Upper Section - - This section is described by the arrow leaving through \(f(x)\). The arrow sweeps from \(f(b)\) to \(f(a)\).
The total area is the combination of these two areas. The results are summarized below.

Horizontal Arrow

lower section

\( a \leq x \leq G(y) \)

arrow leaves through \(G(y)\) and enters through \(x=a\)

\( g(a) \leq y \leq g(b) \)

arrow sweeps from bottom (\(y=g(a)\)) to top (\(y=g(b)\))

upper section

\( a \leq x \leq F(y) \)

arrow leaves through \(F(y)\) and enters through \(x=a\)

\( f(b) \leq y \leq f(a) \)

arrow sweeps from bottom (\(y=f(b)\)) to top (\(y=f(a)\))

Type 1 and Type 2 Regions

Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.

Here is a quick video clip going into more detail on Type 1 and Type 2 regions.

Krista King Math - type I and type 2 regions [1min-39secs]

video by Krista King Math

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, describe the areas defined by the equations both vertically and horizontally. For some videos, the practice problem solutions may show other topics including integrals, some double or triple, which do not impact your solution and you can safely ignore if you are not studying or have not learned them yet.

\(y=2x, y=x^2\)

Problem Statement

\(y=2x, y=x^2\)

Final Answer

vertically \(\{(x,y):~0\leq x\leq2,~x^2\leq y\leq2x\}\)
horizontally \(\{(x,y):~0\leq y\leq4,~y/2\leq x\leq\sqrt{y}\}\)

Problem Statement

\(y=2x, y=x^2\)

Solution

749 solution video

video by Dr Chris Tisdell

Final Answer

vertically \(\{(x,y):~0\leq x\leq2,~x^2\leq y\leq2x\}\)
horizontally \(\{(x,y):~0\leq y\leq4,~y/2\leq x\leq\sqrt{y}\}\)

close solution

\(x=1, y=x\) and the x-axis

Problem Statement

\(x=1, y=x\) and the x-axis

Final Answer

vertically \(\{(x,y):~0\leq x\leq1,~0\leq y\leq x\}\)
horizontally \(\{(x,y):~0\leq y\leq1,~y\leq x\leq1\}\)

Problem Statement

\(x=1, y=x\) and the x-axis

Solution

750 solution video

video by Dr Chris Tisdell

Final Answer

vertically \(\{(x,y):~0\leq x\leq1,~0\leq y\leq x\}\)
horizontally \(\{(x,y):~0\leq y\leq1,~y\leq x\leq1\}\)

close solution
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