Question

Use double integrals to find the indicated volumes. Above the triangle with vertices \((0,0),(2,0),\) and \((2,1),\) and below the paraboloid \(z=24-x^{2}-y^{2}\).

Short answer

The volume is \(\frac{131}{6} \).

Step-by-step solution

Identify and Sketch the Region of Integration

The given region is a triangle with vertices at \(0,0\), \(2,0\), and \(2,1\). Sketch this triangle in the \(x-y\) plane to visualize the area over which we will integrate.

Set Up the Double Integral

The volume V under the paraboloid \(z=24-x^{2}-y^{2}\) and above the triangular region can be represented by the double integral: \[ V = \iint_R (24 - x^{2} - y^{2}) \, dA \] where R is the triangular region.

Determine the Limits of Integration

To perform the integration, determine the limits for \(x\) and \(y\). The region R can be described as follows:- \(x\) ranges from 0 to 2.- For a fixed value of \(x\), \(y\) ranges from 0 to \((1/2)x\).Thus, the limits of integration are \[\int_{0}^{2} \int_{0}^{\frac{x}{2}} (24 - x^2 - y^2) dy \, dx\]

Evaluate the Inner Integral

First, integrate with respect to \(y\): \[\int_{0}^{\frac{x}{2}} (24 - x^2 - y^2) \, dy \] Compute this integral: \[\int_{0}^{\frac{x}{2}} (24 - x^2 - y^2) \, dy = \left[ 24y - x^2y - \frac{y^3}{3} \right]_0^{\frac{x}{2}}\] Evaluate this at the bounds: \[24\left(\frac{x}{2}\right) - x^2\left(\frac{x}{2}\right) - \frac{\left(\frac{x}{2}\right)^3}{3} - (0) = 12x - \frac{x^3}{2} - \frac{x^3}{24} \] Simplify the expression: \[12x - \frac{x^3}{2} - \frac{x^3}{24} = 12x - \frac{13x^3}{24}\]

Evaluate the Outer Integral

Now, integrate the resulting expression with respect to \(x\): \[\int_{0}^{2} \left(12x - \frac{13x^3}{24}\right) \, dx\] Compute this integral: \[ \left[ 6x^2 - \frac{13x^4}{96} \right]_0^{2} \] Evaluate this at the bounds: \[\left(6(2)^2 - \frac{13(2)^4}{96}\right) - \left(6(0)^2 - \frac{13(0)^4}{96}\right) = (24 - \frac{208}{96}) - 0 \] Simplify the expression: \[24 - \frac{13}{6} = \frac{144}{6} - \frac{13}{6} = \frac{131}{6}\] Therefore, \[V = \frac{131}{6}\]

Key concepts

Volume Calculation

Calculating the volume under a surface and above a region in the plane is a common application of double integrals. In this exercise, we find the volume under a paraboloid and above a triangular region in the x-y plane. By setting up and evaluating a double integral, we can determine this volume precisely. We start by defining the region based on given vertices, then write the integral with the appropriate limits, and finally perform the integration to get the volume.

Paraboloid

A paraboloid is a surface in three-dimensional space defined by an equation of the form:

This shape is like a 3D parabola. In our exercise, the paraboloid equation is: represents the height (z-value) at any point (x, y) in the region. To find the volume under this surface and above the specified region, we use a double integral where the integrand is the function representing the paraboloid.

Region of Integration

The region of integration is the area over which we'll integrate. It's important to visualize this region to set up the correct limits for our integrals. In this exercise, the region is a triangle with vertices at (0, 0), (2, 0), and (2, 1). When sketching this triangle in the x-y plane:

• Ensure you clearly define the boundaries of the region for accurate limits of integration.
  To describe the region more accurately, we identify the limits for each variable within the triangular area.

Limits of Integration

The limits of integration define the range of values over which we will integrate for x and y. In this problem, we first:

• Determine the x-limits: x ranges from 0 to 2.
• For a fixed value of x, y ranges from 0 to (1/2)x.

Thus, our double integral becomes:
Next, we set up the outer integral with respect to x, and for each x, we set up the inner integral with respect to y. This way, we ensure we cover the entire triangular region.

Integral Evaluation

To find the volume, we need to evaluate the double integral. We do this in stages:

• Evaluating the inner integral:
We integrate the function with respect to y first, keeping x constant. For our specific problem, this means integrating:
Here's the result after integration:

• Evaluating the outer integral:
The next step is to integrate the resulting expression with respect to x:

We then evaluate this at the bounds, simplify the expression, and ultimately find the volume, which in this case is:
Mastery of these steps is crucial in successfully using double integrals for volume calculations.