91Ó°ÊÓ

To understand how to find the volume under a surface, imagine a 3D space where a surface is described by a function, say, \(z = f(x, y)\). We often want to know the volume of the region below this surface and above a specific area on the \(xy\)-plane. This volume can be found using double integrals.

For our specific function \(f(x, y) = e^{y^2}\), the volume under this surface can be visualized as the accumulated 'heights' given by \(e^{y^2}\) across a certain region. To find this volume accurately, we use a double integral that sums up these heights over the entire given region.

Identifying the region of integration is a crucial step when setting up a double integral. This region is basically the area in the \(xy\)-plane over which we'll integrate our function.

In our example, the region \(R\) is bounded by the curves \(x=2y\), \(x=0\), and \(y=1\). To describe this region clearly:

• the line \(x=2y\) suggests that \(x\) values stretch from 0 to \(2y\) for each \(y\) value,
• the line \(x=0\) is simply the y-axis, indicating that our region starts from \(x=0\) for each \(y\),
• and the horizontal line \(y=1\) indicates that \(y\) ranges from 0 to 1.

Hence, when we set up our double integral, we integrate \(x\) from 0 to \(2y\) and \(y\) from 0 to 1.

Sometimes, solving an integral directly can be tough, and substitution can help simplify it. Substitution in integration involves transforming the integral into a different variable to make it easier to solve.

In our example, after performing the first integral with respect to \(x\), we are left with the integral \(\int_0^1 2y e^{y^2} \, dy\). To simplify this, we use the substitution method: let \(u = y^2\). Then, the differential \(du = 2y \, dy\). Notice how this transforms our integral’s variables:

• When \(y=0\), \(u\) also equals 0.
• When \(y=1\), \(u\) equals 1.

This changes the integral to \( \int_0^1 e^u \, du\). Solving this integral is more straightforward:

\[\begin{array}{l} \int_0^1 e^u \, du = e^u \Big|_0^1 = e - 1 \ \end{array} \]
This substitution method thus simplifies our calculations significantly!