Calculating Areas Bounded by the y-Axis

# Calculating Areas Bounded by the y-Axis

Suppose that we want to calculate the area trapped between the curve $f(y) = x$ and the $y$-axis. Recall that when we were trying to calculate areas trapped between a curve and the $x$-axis, we used the formula $\mathbf{Area} = \int_a^b f(x) \: dx$ (provided $f$ lies above the $x$-axis on $[a, b]$. However, we can swap the variables out to get the following formula:

(1)
\begin{align} \int_a^b f(y) \: dy \end{align}

## Example 1

**Calculate the area trapped between the function $f(y) = -y^2 + 2$ and the y-axis.**

Let's first find where the curve $f(y)$ intersects the $y$-axis. This will be our upper and lower bounds of integration.

(2)
\begin{align} f(y) = -y^2 + 2 \\ 0 = -y^2 + 2 \\ 2 = y^2 \\ y = \sqrt(2), -\sqrt(2) \end{align}

The following graph represents the area we intend to find:

We can now integrate using the formula from above.

(3)
\begin{align} \mathbf{Area} = \int_{-\sqrt{2}}^{\sqrt{2}} (-y^2 + 2) \: dy \\ \mathbf{Area} = [-\frac{y^3}{3} + 2y] \bigg |_{-\sqrt{2}}^{\sqrt{2}} \\ \mathbf{Area} = [-\frac{y^3}{3} + 2y] \bigg |_{-\sqrt{2}}^{\sqrt{2}} \\ \mathbf{Area} \approx 3.7712 \end{align}

## Example 2

**Calculate the area bounded by the curve $x = \sin y + 1$, $y = 0$, $y = \pi$ and the $y$-axis.**

We note that our lower bound $a = 0$, while our upper bound $b = \pi$, and therefore:

(4)
\begin{align} \mathbf{Area} = \int_{0}^{\pi} \sin y + 1 \: dy \\ \mathbf{Area} = \left [ -\cos y + y \right ]_{0}^{\pi} \\ \mathbf{Area} = (-(-1) + \pi) - (-(1) + 0) \\ \mathbf{Area} = \pi + 2 \end{align}

## Example 3

**Calculate the area bounded by the curve $x = \sqrt{4 + y^2}$, $y = -4$, $y = 4$, and the $y$-axis.**

We note that our lower bound $a = -4$ and our upper bound $b = 4$. Therefore:

(5)
\begin{align} \mathbf{Area} = \int_{-4}^{4} \sqrt{4 + y^2} \: dy \end{align}

We will now have to use a trigonometric substitution. Let $y = 2 \tan \theta$ so that $dy = 2 \sec ^2 \theta \: d \theta$. Making this substitution we get that:

(6)
\begin{align} \mathbf{Area} = \int_{\alpha}^{\beta} \sqrt{4 + (2\tan \theta)^2} \cdot 2 \sec ^2 \theta \: d \theta \\ \mathbf{Area} = \int_{\alpha}^{\beta} \sqrt{4 + 4 \tan ^2 \theta} \cdot 2 \sec ^2 \theta \: d \theta \\ \mathbf{Area} = \int_{\alpha}^{\beta} \sqrt{4(1 + \tan ^2 \theta)} \cdot 2 \sec ^2 \theta \: d \theta \\ \mathbf{Area} = \int_{\alpha}^{\beta} \sqrt{4\sec ^2 \theta} \cdot 2 \sec ^2 \theta \: d \theta \\ \mathbf{Area} = \int_{\alpha}^{\beta} \sqrt{4\sec ^2 \theta} \cdot 2 \sec ^2 \theta \: d \theta \\ \mathbf{Area} = \int_{\alpha}^{\beta} 2 \sec \theta \cdot 2 \sec ^2 \theta \: d \theta \\ \mathbf{Area} = 4 \int_{\alpha}^{\beta} \sec \theta (1 + \tan^2 \theta) d \theta \\ \mathbf{Area} = 4 \int_{\alpha}^{\beta} \sec \theta + \sec \theta \tan^2 \theta d \theta \\ \end{align}

We will not continue the example further as it is rather tedious, that $A = 23.66$.