Question

Set up a double integral to find the volume of the solid bounded by the graphs of the equations. \(z=0, z=x^{2}, x=0, x=2, y=0, y=4\)

Short answer

The volume of the solid is \(\frac{32}{3}\) cubic units.

Step-by-step solution

Identify the Limits of Integration

The limits of integration would be the given x and y boundaries. Here, \(x\) varies from 0 to 2 and \(y\) varies from 0 to 4.

Set up the Double Integral

The double integral would be set up to find volume \(V\) under the function \(z=x^{2}\) with respect to \(x\) and \(y\). With the given \(x\) and \(y\) limits, the double integral becomes: \[V = \int_{0}^{4} \int_{0}^{2} (x^{2}) dx\, dy\]

Evaluate the Inner Integral

The inner integral with respect to \(x\) is calculated first, evaluating the antiderivative of \(x^{2}\) at the limits of \(x\) (from 0 to 2). This gives, \[ V = \int_{0}^{4} [\frac{1}{3} x^{3} |_{0}^{2}] dy = \int_{0}^{4} [\frac{8}{3} - 0] dy = \int_{0}^{4} \frac{8}{3} dy\]

Evaluate the Outer Integral

The second integral is then evaluated at its limits \(y=0\) to \(y=4\), giving us\[ V = [\frac{8}{3} y |_{0}^{4}] = \frac{32}{3} - 0 = \frac{32}{3}\]