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# Calculus

## Finding the Area between Two Curves using Vertical Elements

To find the area between two curves, we will follow the process of using the length of the rectangle multiplied with the width and adding them all up, which is similar to finding the area under a curve. However, there is one difference: the length of the rectangle is no longer the distance from the curve to the axis. Now, the length of the rectangle is the distance from the lower curve to the upper curve. In general, the area between 2 curves, $$f(x)$$ and $$g(x)$$, is given by evaluating the equation:

$$A=\int_a^b(f(x)-g(x)) \ dx$$

Example

Take these graphs for the curves $$y=\sqrt{x}$$ and $$y=x^2$$:

From the first graph, the length of the rectangles from the axis to curve, at any point, is $$y=\sqrt{x}$$. From the second graph, the length of the rectangles from the axis to curve, at any point, is $$y=x^2$$. However, to find the area in between the curves, we need to know the length of the rectangles in between the curves. To find this, we need to find the difference between them, by subtracting the lower curve from the upper curve:

So the length of the rectangles in between the curves, at any point, is $$length=\sqrt{x}-x^2$$. The width of the rectangles is $$dx$$, as in the example from

The calculations for this example would be as follows:

$$Area=\int_0^1(\sqrt{x}-x^2) \ dx$$

$$=(\frac{2}{3}x^{\frac{3}{2}}-\frac{x^3}{3})|_{x=0}^{x=1}$$

$$=\frac{2}{3}(1)^{\frac{3}{2}}-\frac{1^3}{3}$$

$$=\frac{2}{3}-\frac{1}{3}$$

$$=\frac{1}{3}$$

## Finding the Area between Two Curves using Horizontal Elements

In the above example, the length of the rectangle was in the vertical direction. But we can also calculate the area between two curves by drawing these rectangles in the horizontal direction. We'll demonstrate this idea below:

Example

Let’s take the case of the 2 equations, $$x=4$$ and $$x=(y-3)^2$$.

If we try to use vertical rectangles again, we run into a problem. What is the length of the rectangles? The top of the rectangle and the bottom of the rectangle both touch the same curve, so we can’t subtract to find the difference.

The solution is that we have to use horizontal rectangles instead. This way, one end of the rectangle is touching one curve, and the other end of the rectangle is touching the other. This way, we can subtract the two graphs to find the difference between them, which is the length of the rectangles in between the curves.

When we were using vertical rectangles, we subtracted the lower curve from the upper curve. Now that we’re using horizontal rectangles, we will subtract the left curve from the right curve. So, in our example case, the length of the rectangles at any point between the curves is $$length=4-(y-3)^2$$.

When we were using vertical rectangles, the width of the rectangle was an infinitely small length along the x-axis, and so we named it $$dx$$. Now, with horizontal rectangles, the width of the rectangle is an infinitely small length along the y-axis, which is named $$dy$$.

So the area of one of these rectangles is $$length \times width = (4-(y-3)^2) \cdot dy$$. To add all of these areas together to find the total area between the curves, we integrate:

$$\int_b^a4-(y-3)^2 \ dy$$

Before we can evaluate this, we need to know the bounds of the integration, $$a$$ and $$b). Since we are using horizontal rectangles, instead of our bounds being on the x-axis, like they were in the previous examples, now our bounds are on the y-axis. Looking at our graph, we can see that the 2 bounds of our integration will be at the points where the lines intersect. To find these points of intersection, we need to substitute one of the variables, and solve for the remaining one: Equation 1: \(x=4$$

Equation 2: $$x=(y-3)^2$$

Since we are looking for the bounds on the y-axis, we know we need to solve for the y variable, and can eliminate x. Therefore:

$$4=(y-3)^2$$

$$y=3 \pm 2$$

$$y=1, y=5$$

Now that we’ve solved for the points of intersection, giving us the bounds of our integration, we can perform the following calculations to determine the area between these two curves:

$$Area=\int_1^5 4-(y-3)^2 \ dy$$

$$=\int_1^5 4-(y^2-6y+9) \ dy$$

$$=(-\frac{y^3}{3}+3y^2-5y)|_{x=1}^{x=5}$$

$$=8.33$$