Questions: The angle at the vertex is π / 3, and the top is flat and at a height of 7√3. Write the limits of integration for ∫W d V in the following coordinates (do not reduce the domain of integration by taking advantage of symmetry): (a) Cartesian: With a=-21, b=21 c=-21, d=21 e=0, and f=7√3 Volume =∫a^b ∫c^d ∫e^f d square d square

Solution Steps

To determine the limits of integration for the given volume in Cartesian coordinates, we need to understand the geometric shape described. The angle at the vertex and the height suggest a conical or pyramidal shape. The limits for \(x\) and \(y\) are given as \([-21, 21]\), and the height \(z\) ranges from \(0\) to \(7\sqrt{3}\). The integration will be performed over these limits to calculate the volume.

To find the volume of the solid defined by the given limits, we set up the triple integral:

\[ V = \int_{-21}^{21} \int_{-21}^{21} \int_{0}^{7\sqrt{3}} dV \]

The volume integral simplifies to:

\[ V = \int_{-21}^{21} \int_{-21}^{21} \int_{0}^{7\sqrt{3}} 1 \, dz \, dy \, dx \]

Calculating the innermost integral with respect to \(z\):

\[ \int_{0}^{7\sqrt{3}} 1 \, dz = 7\sqrt{3} \]

Now substituting this result into the double integral:

\[ V = \int_{-21}^{21} \int_{-21}^{21} 7\sqrt{3} \, dy \, dx \]

Next, we evaluate the double integral:

\[ V = 7\sqrt{3} \int_{-21}^{21} \int_{-21}^{21} 1 \, dy \, dx \]

Calculating the area of the \(xy\)-plane:

\[ \int_{-21}^{21} \int_{-21}^{21} 1 \, dy \, dx = (21 - (-21))(21 - (-21)) = 42 \times 42 = 1764 \]

Thus, the volume becomes:

\[ V = 7\sqrt{3} \times 1764 \]

Calculating the final volume:

\[ V = 12348\sqrt{3} \]

Final Answer