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When calculating the volume of a solid formed by revolving a plane region around an axis, we use the concept of solids of revolution. In this exercise, we focus on revolving a region around the y-axis using the shell method. The solid of revolution is obtained when the plane region bounded by the given equations is rotated about the y-axis. This results in a three-dimensional shape, and our goal is to determine its volume. The shell method is particularly useful for these types of problems. It involves considering thin cylindrical shells. By summing up the volumes of these infinitesimally thin shells, we acquire the total volume of the solid. This method is particularly beneficial when dealing with functions expressed in terms of y, or when the region is bounded by the y-axis.

Integrating functions is a fundamental skill in calculus, and various techniques can simplify this process. In our problem, the shell method leads to an integral that represents the volume of the solid. This integral is initially set up using the function's expression related to the height and the radius of the shells. To simplify this integration, it's essential to break down the function into more manageable forms. Here, we multiply the function by the expression for the radius (which is simply 'x' in this case). Then, we can cancel constants outside the integral, allowing us to deal purely with the essential parts of the function that affect its evaluation. These strategies are crucial in making integrals more approachable. By simplifying first, you reduce complexities in calculations, making it easier to apply subsequent techniques, such as substitution, that will further ease the process of finding the solution.

U-substitution is a powerful technique used to simplify the integration of expressions that are difficult to handle in their current form. In this case, the integral of the function was initially complex due to the exponential function combined with 'x'.By utilizing substitution, our aim is to transform the integral into one with a more straightforward antiderivative. We let the exponent of the exponential function become 'u', i.e., set \( u = x^2/2 \), which results in \( du = x \, dx \). This substitution aligns perfectly with another part of the integrand (the 'x' term), allowing the expression to rewrite the integral entirely in terms of 'u'.This process simplifies the calculation of the antiderivative significantly, turning a potentially daunting problem into one that is manageable and standard, greatly facilitating both the integration and subsequent evaluation.

Definite integrals are a key tool in calculating exact values, such as volumes, areas, or other characteristics, from continuous functions. In this scenario, once the integral is set up and simplified through techniques like u-substitution, what's left is evaluating the definite integral. The bounds of the definite integral, from 0 to 1 initially, transform according to the substitution's new variable. In this case, the bounds become corresponding values in terms of 'u', namely from 0 to 0.5. By evaluating the definite integral, we determine the exact volume of the solid formed by rotation. It's important to remember that subtraction is part of this evaluation, calculating the value of the function at the upper limit of integration minus the value at the lower limit, which ultimately gives the solid's volume. The calculation concludes with straightforward arithmetic, taking into account the physical interpretation—ensuring our values align with realistic scenarios, hence the removal of a negative sign in the final step, since physical volumes cannot be negative.