Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the solid whose volume is the indicated iterated integral. $$ \int_{0}^{2} \int_{0}^{2}\left(x^{2}+y^{2}\right) d y d x $$

Step-by-Step Solution

Answer

Sketch a paraboloid over a 2x2 square on the xy-plane.

Step 1: Set Up the Problem

Begin by understanding that the integral is set up in the order \( \int_0^2 \int_0^2(x^2 + y^2)\,dy\,dx\). This implies that we have a region in the xy-plane where x and y range from 0 to 2 over a surface described by \(z = x^2 + y^2\) within these bounds.

Step 2: Define the Region in the XY-plane

The region of integration in the xy-plane is defined by the rectangle with x ranging from 0 to 2 and y ranging from 0 to 2. Visualize this as a square on the xy-plane.

Step 3: Understand the Surface Equation

The equation of the surface is \(z = x^2 + y^2\). This represents a paraboloid opening upwards, but as z is explicitly dependent on x and y, for each point (x,y) in the defined region, there is a corresponding z-value on this surface.

Step 4: Visualize the Solid

The solid whose volume the integral represents is the volume under the surface \(z = x^2 + y^2\) and over the square region, where both x and y range from 0 to 2. Sketch a square on the xy-plane with the described surface rising from it.

Step 5: Sketch the Solid

Begin your sketch with the axes. Draw the square region on the xy-plane first. Then, sketch the paraboloid that starts at the origin and increases as both x and y increase up to 2. The sections of the solid closer to the xy-plane will have lower z-values (like near the origin), and as you move towards (2,2), z increases.

Key Concepts

Volume of Solids

When we talk about the volume of solids in the context of iterated integrals, we are usually concerned with finding the space occupied below a surface and above a certain region in the xy-plane. This is exactly what the given integral does—it calculates the volume under the surface \(z = x^2 + y^2\) over the region from \(x = 0\) to \(x = 2\) and \(y = 0\) to \(y = 2\).

This approach is very useful because it breaks down the problem into more manageable bite-sized sections. The iterated integral first computes tiny volume elements over strips of the region, adding them all up to give the total volume. Here, the iterated integral allows us to sum up the infinitesimal change \(dy\) first, followed by \(dx\), within the prescribed bounds, thus giving the total volume of the solid over the defined region.

Paraboloid Surface

The surface described by the equation \(z = x^2 + y^2\) is a type of paraboloid. Specifically, it is an elliptic paraboloid, where the cross-sections parallel to the xy-plane are circles, and those vertical to it are parabolas. This surface opens upwards, meaning that as you move away from the origin in the xy-plane, the z-value gets larger.

In the context of this problem, the paraboloid acts as the ceiling or boundary above the square region in the xy-plane. This understanding is crucial for solving the iterated integral because the z-value on this surface gives the height at any \