91Ó°ÊÓ

An iterated integral is a way to compute a multiple integral, one integral at a time. This approach is particularly useful in multivariable calculus when determining the volume of a solid region under a curve or surface. It simplifies calculations by breaking down a complex multidimensional integral into simpler single-variable integrals.
To compute an iterated integral, choose an order of integration — typically integrating with respect to one variable while treating others as constants. In our case, we integrate first with respect to \(y\), then \(x\). This step-by-step method works through nested layers of integration, effectively capturing the contribution of each variable across the volume.
For the given solid:

• First, integrate with respect to \(y\), treating \(x\) as constant over the interval determined by the projection of the triangle in the \(xy\)-plane.
• Next, integrate the obtained expression with respect to \(x\) over its limits.

Each integration unwinds a layer, finally computing the volume under the paraboloid and above the triangular region.

A paraboloid is a 3-dimensional geometric surface, shaped somewhat like a stretched parabola. It has important applications in physics and engineering, often representing the path of light or physical objects under certain conditions.
The given paraboloid is expressed by the equation \(z = 25 - x^2 - y^2\). This is a standard elliptic paraboloid which curves downward, as defined by the negative coefficients of \(x^2\) and \(y^2\).

• At the peak or the vertex of this paraboloid, \(x = 0\) and \(y = 0\), making \(z = 25\).
• As \(x\) and \(y\) move away from zero, \(z\) decreases, forming a dome-like shape.

By integrating under this downward-curving surface over a specific region, we calculate the enclosed volume beneath it.

Region projection refers to mapping a 3D shape onto a 2D plane, helping to understand the limits of integration. In our case, we deal with a triangular region in the \(xy\)-plane, where the solid's 'shadow' is cast under the paraboloid. This helps visualize how the function extends over a specific area, guiding the setup of our iterated integral.
The triangle's vertices are

• \((-3, -4)\)
• \((-3, 4)\)
• \((5, 0)\)

Using these points, determine line equations to define the triangle:

• The vertical line for the side from \((-3, -4)\) to \((-3, 4)\): \(x = -3\).
• The sloped line from \((-3, 4)\) to \((5, 0)\): \(y = -\frac{1}{2}x + \frac{7}{2}\).
• The line between \((5, 0)\) and \((-3, -4)\): \(y = \frac{1}{2}x - \frac{5}{2}\).

These equations define the integration limits for \(y\) at any given \(x\), ensuring the integration correctly maps the entire base of the solid.

Volume calculation involves integrating a surface or solid across a defined region. For a complex shape like a paraboloid projected onto a triangular base, it requires setting up and evaluating an iterated integral that combines geometrical understanding with calculus.
Here's how it's done:

• Set integration limits for \(x\) from \(-3\) to \(5\), spanning the triangle's width.
• For each \(x\), let \(y\) vary between the lines \(y = \frac{1}{2}x - \frac{5}{2}\) and \(y = -\frac{1}{2}x + \frac{7}{2}\).

With these limits, solve the iterated integral to find the volume:\[\int_{-3}^{5} \int_{\frac{1}{2}x - \frac{5}{2}}^{-\frac{1}{2}x + \frac{7}{2}} (25 - x^2 - y^2) \, dy \, dx\] By evaluating this, you capture the total space the solid occupies between the paraboloid above and the triangular base, obtaining the volume of the solid.