Question

Evaluate the double integral. $$ \int_{0}^{2} \int_{0}^{6 x^{2}} x^{3} d y d x $$

Short answer

The value of the given double integral is 64.

Step-by-step solution

- Simplification of Inner Integral

To evaluate the double integral, start by evaluating the inner integral. The integration to be carried out is over \(y\) and can be written as \( \int_{0}^{6x^{2}} x^{3} dy \), where \(x^3\) is a constant with respect to \(y\). The integral of a constant over a range {a, b} is simply the constant multiplied by the length of the interval. So this gives \( x^{3} * (6x^{2} - 0) = 6x^{5} \).

- Calculation of Outer Integral

The result from the first step \(6x^{5}\) is then to be integrated over \(x\), from 0 to 2. This yields \( \int_{0}^{2} 6x^{5} dx \). Using the power rule for integration, increase the power of \(x\) by 1 and divide by the new power, to get \( \frac{6}{6} x^{6} \) - this simplifies to \(x^{6}\). Evaluate this from 0 to 2 to arrive at \(2^{6} - 0^{6} = 64\).

- Interpretation of Result

The result 64 tells you the result of the given double integral. In the context of multivariable calculus, double integrals can be interpreted as the volume under a surface over a given region. Hence, the given function \(x^{3}\) and the given bounds would create a surface in three dimensions, and the double integral, 64, is the volume below this surface and above the plane of the region defined by \(x = [0, 2]\) and \(y = [0, 6x^{2}]\).