Problem 25 - College Algebra and Calculus: An Applied Approach

Question

Use a double integral to find the volume of the solid bounded by the graphs of the equations. $$ z=x^{2}, z=0, x=0, x=2, y=0, y=4 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Geometrical Meaning

The equations \(z=x^{2}, z=0, x=0, x=2, y=0, y=4\) define a solid that is essentially a 3D region under the curve \(z=x^{2}\) between \(x=0\) and \(x=2\), and \(y=0\) and \(y=4\). The solid is bounded below by \(z=0\) (the xy-plane).

Step 2: Set Up Double Integral

The volume of a solid bounded by a surface \(z=f(x,y)\) and plane \(z=0\) can be calculated through the double integral \(\int_Y \int_X f(x,y)dxdy\). Now, in this case, \(f(x,y)=x^{2}\). Also, \( 0 \leq x \leq 2\) and \(0 \leq y \leq 4\). So, the double integral becomes \(\int_{0}^{4} \int_{0}^{2} x^{2} dx dy\).

Step 3: Compute Inner Integral

The inner integral concentrates on the variable \(x\). It can be computed by using standard integral rules, leading to \(\int_{0}^{2} x^{2} dx = \left[ \frac{1}{3}x^{3} \right]_{0}^{2} = \frac{8}{3}\). The resulting (simplified) integral is \(\int_{0}^{4} \frac{8}{3} dy \).

Step 4: Compute Outer Integral

Now, let's compute the outer integral which is with respect to the variable \(y\). This results in \(\int_{0}^{4} \frac{8}{3} dy = \left[ \frac{8}{3}y \right]_{0}^{4} = \frac{32}{3}\).