Calculating Areas Bounded by Curves

We all know that integrals give areas; i.e. definite integrals given areas under curves over a particular interval. Now, let us ask a different question. If we had two curves and we took the region bounded between those two curves, what would be the area? Let us start with an example.

Let us say we want to find the bounded area between the graphs of $y = x^2$ and $y = \sqrt {x}$. We can clearly see the lemon-shaped region below.

The easiest way to think about it is the following. Let's say we wanted to define $\mathcal{R}$ to be the entire region in the first quadrant above the graph $y=x^2$. Now define the entire region below the graph of $y=\sqrt{x}$ to be $\mathcal{S}$. It is clear that the region $\mathcal{R} \cap \mathcal{S}$ is equal to the area in the first quadrant below $y = \sqrt {x}$ minus the area under the curve $y = x^2$. Try to see why this is and once you are done, click Continue.

Now that we have understood what type of problem we are solving, we can begin converting to math; namely, integrals. Note that the area we are interested in runs between the two intersections of the curves.

Therefore, we are ready to convert into integrals. The area under the curve of $y = \sqrt {x}$ in the given interval is $$\displaystyle\int_0^1 \sqrt {x} \, \mathrm{d}x$$ and the area under the curve of $y = x^2$ in the given interval is $$\displaystyle\int_0^1 x^2 \, \mathrm{d}x.$$ Now think about how to finish and then continue.

Therefore, the answer is just $$\displaystyle\int_0^1 \sqrt {x} \, \mathrm{d}x - \displaystyle\int_0^1 x^2 \, \mathrm{d}x = \displaystyle\int_0^1 \left( \sqrt {x} - x^2 \right) \, \mathrm{d}x.$$

Yes, we find that $$\begin {align*} \displaystyle\int_0^1 \left( \sqrt {x} - x^2 \right) \, \mathrm{d}x &= \left[ \frac {2}{3} x^{\tfrac{3}{2}} - \frac {x^3}{3} \right]_0^1 \\&= \left( \frac {2}{3} - \frac {1}{3} \right) - \left( 0 - 0 \right) \\&= \boxed {\frac{1}{3}}. \end {align*}$$

Great, we have solved the problem! Think about how you might generalize this.

The answer turns out to be that we can subtract just like we did with the two functions above and we will get our answer over a given interval.

$\implies$ Final Result. If $f(x)$ and $g(x)$ represent two curves in the $xy$-plane, the area bounded by these two curves between $x=a$ and $x=b$ is represented by $$\displaystyle\int_a^b \left| f(x) - g(x) \right| \, \mathrm{d}x$$

Exercise. Verify the above result.

Note. Note that the same works if we rotate the plane and we write everything in terms of $y$. We would simply have to change the differential to $\mathrm{d}y$ instead of $\mathrm{d}x$.

We already know that an integral is an area under a curve. An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

(Video) Finding Areas Between Curves

by PatrickJMT

Here's a video by patrickJMT where he shows you how to find any area that's between two curves.

For example, suppose you had some function $f(x)$.

And another function below it, $g(x)$, and both functions are defined from $a$ to $b$.

How would you find the area that is between them? Well, we can use integrals! We set the integral from $a$ to $b$ and then integrate the top curve minus the bottom curve. $$\int_a^b\left[f(x)-g(x)\right]dx$$

Let's look at some examples. In the first, the curves are given to us.

We want to calculate the area between the two curves from $(0,0)$ to $(6,12)$.

We start by integrating from the smallest $x$-coordinate to the largest $x$-coordinate, i.e. from 0 to 6. $$\int_0^6$$

Next, we want to take the top curve and subtract the bottom curve. On our interval the line $y=2x$ is our top curve (because it's higher up) and $x^2-4x$ is our bottom curve. $$\int_0^6\left[2x-(x^2-4x)\right]dx$$

Now that we set the problem up, we solve. $$\begin{aligned} &\int_0^6\left[2x-(x^2-4x)\right]dx\\\ &\int_0^6\left[2x-x^2+4x\right]dx\\\ &\int_0^6\left[6x-x^2\right]dx\\\ &\int_0^6 6x\text{ }dx-\int_0^6 x^2\text{ }dx\\\ &6\int_0^6 x\text{ }dx-\int_0^6 x^2\text{ }dx\\\ &6\left[\frac12x^2\right]_0^6-\left[\frac13x^3\right]_0^6\\\ &3\left[x^2\right]_0^6-\frac13\left[x^3\right]_0^6\\\ &3\left[(6)^2-(0)^2\right]-\frac13\left[(6)^3-(0)^3\right]\\\ &3\left[36-0\right]-\frac13\left[216-0\right]\\\ &3(36)-\frac{216}{3}\\\ &\boxed{36} \end{aligned}$$

In this second example we're given the equations and have to determine which is the top curve and which is the bottom curve.

$y=8-x^2$, $y=x^2$, $x=-3$, $x=3$

Start by graphing each curve.

We can set the two curves equal to each other to find their points of intersection. $$\begin{aligned} 8-x^2&=x^2\\ 8&=2x^2\\ 4&=x^2\\ \pm2&=x \end{aligned}$$

From left to right the three sections of integrals are from $-3$ to $-2$, $-2$ to $2$, and $2$ to $3$. $$\int_{-3}^{-2}\left[x^2-(8-x^2)\right]dx+\int_{-2}^{2}\left[(8-x^2)-x^2\right]dx+\int_{2}^{3}\left[x^2-(8-x^2)\right]dx$$

Now that the equation is set up we can solve the integrals. Let's start with the first. $$\begin{aligned} &\int_{-3}^{-2}\left[x^2-(8-x^2)\right]dx\\\ &\int_{-3}^{-2}\left[x^2-8+x^2\right]dx\\\ &\int_{-3}^{-2}\left[2x^2-8\right]dx\\\ &\int_{-3}^{-2} 2x^2\text{ }dx-\int_{-3}^{-2} 8\text{ }dx\\\ &2\int_{-3}^{-2} x^2\text{ }dx-8\int_{-3}^{-2} \text{ }dx\\\ &2\left[\frac13x^3\right]_{-3}^{-2}-8\left[x\right]_{-3}^{-2}\\\ &\frac23\left[x^3\right]_{-3}^{-2}-8\left[x\right]_{-3}^{-2}\\\ &\frac23\left[(-2)^3-(-3)^3\right]-8\left[(-2)-(-3)\right]\\\ &\frac23(19)-8(1)\\\ &\frac{14}3 \end{aligned}$$

Next, the second. $$\begin{aligned} &\int_{-2}^2\left[(8-x^2)-x^2\right]dx\\\ &\int_{-2}^2\left[8-x^2-x^2\right]dx\\\ &\int_{-2}^2\left[8-2x^2\right]dx\\\ &\int_{-2}^2 8\text{ }dx-\int_{-2}^2 2x^2\text{ }dx\\\ &8\int_{-2}^2 dx-2\int_{-2}^2 x^2\text{ }dx\\\ &8\left[x\right]_{-2}^2-2\left[\frac13x^3\right]_{-2}^2\\\ &8\left[x\right]_{-2}^2-\frac23\left[x^3\right]_{-2}^2\\\ &8\left[(2)-(-2)\right]-\frac23\left[(2)^3-(-2)^3\right]\\\ &8(4)-\frac23(16)\\\ &\frac{64}3 \end{aligned}$$

Finally, the last integral. $$\begin{aligned} &\int_{2}^{3}\left[x^2-(8-x^2)\right]dx\\\ &\int_{2}^{3}\left[x^2-8+x^2\right]dx\\\ &\int_{2}^{3}\left[2x^2-8\right]dx\\\ &\int_{2}^{3} 2x^2\text{ }dx-\int_{-3}^{-2} 8\text{ }dx\\\ &2\int_{2}^{3} x^2\text{ }dx-8\int_{-3}^{-2} \text{ }dx\\\ &2\left[\frac13x^3\right]_{2}^{3}-8\left[x\right]_{2}^{3}\\\ &\frac23\left[x^3\right]_{2}^{3}-8\left[x\right]_{2}^{3}\\\ &\frac23\left[(3)^3-(2)^3\right]-8\left[(3)-(2)\right]\\\ &\frac23(19)-8(1)\\\ &\frac{14}3 \end{aligned}$$

Combining these three we get: $$\frac{14}3+\frac{64}3+\frac{14}3=\frac{92}3$$

(Videos) Area Between Curves with Respect to $y$

by PatrickJMT

In these videos by patrickJMT he shows you how to find the area between curves by taking the integral with respect to $y$. Previously it was with respect to $x$. It will be roughly the same process. The integral bounds will be the min and max of the $y$-axis as opposed to the $x$-axis.

Suppose you have two functions $f(y)$ and $g(y)$ and want to find the area between the two curves.

We want to find the $y$-coordinates of their points of intersection, i.e. $c$ and $d$. Then you take the integral of the right most function minus the left most function. $$\int_c^d\left[f(y)-g(y)\right]dy$$

The example he looks at is finding the area between the curves $x=y^3-y$ and $x=1-y^4$.

The first step is to graph the curves.

Now set the curves equal to each other in order to see where they intersect. $$\begin{aligned} y^3-y&=1-y^4\\ y^4+y^3-y-1&=0\\ (y^4+y^3)-(y+1)&=0\\ y^3(y+1)-(y+1)&=0\\ (y^3-1)(y+1)&=0\\ y&=\pm1 \end{aligned}$$

We need to integrate from -1 to 1 and integrate the right most curve, $1-y^4$, minus the left most curve, $y^3-y$. $$\int_{-1}^1\left[(1-y^4)-(y^3-y)\right]dy$$

Finally, integrate. $$\begin{aligned} &\int_{-1}^1\left[(1-y^4)-(y^3-y)\right]dy\\\ &\int_{-1}^1\left[1-y^4-y^3+y\right]dy\\\ &\int_{-1}^11\text{ }dy-\int_{-1}^1y^4\text{ }dy-\int_{-1}^1y^3\text{ }dy+\int_{-1}^1y\text{ }dy\\\ &\left[y\right]_{-1}^1-\left[\frac15y^5\right]_{-1}^1-\left[\frac14y^4\right]_{-1}^1+\left[\frac12y^2\right]_{-1}^1\\\ &\left[y\right]_{-1}^1-\frac15\left[y^5\right]_{-1}^1-\frac14\left[y^4\right]_{-1}^1+\frac12\left[y^2\right]_{-1}^1\\\ &\left[1-(-1)\right]-\frac15\left[(1)^5-(-1)^5\right]-\frac14\left[(1)^4-(-1)^4\right]+\frac12\left[(1)^2-(-1)^2\right]\\\ &2-\frac25-0+0\\\ &\boxed{\frac85} \end{aligned}$$