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Chapter 12: Problem 7

Set up a triple integral for the volume of the solid. The solid in the first octant bounded by the coordinate planes and the plane \(z=4-x-y\)

Short Answer

Expert verified
The triple integral for the volume of the solid bounded by the plane \(z=4-x-y\) and the first octant is \(\int_0^4 \int_0^{4-x} \int_0^{4-x-y} dz dy dx\).

Step by step solution

01

Identify the Ranges for x, y, and z

The plane \(z=4-x-y\) intercepts the x-axis when y and z are 0 Hence, x ranges from 0 to 4. The plane intercepts the y-axis when x and z are 0. Hence, y ranges from 0 to 4-x. The z varies from 0 (the xy-plane) up to the plane \(z=4-x-y\).
02

Setting Up The Integral

The volume of the solid in the first octant bounded by the coordinate planes and the given plane can be represented by the triple integral \(\int_0^4 \int_0^{4-x} \int_0^{4-x-y} dz dy dx\). Each of the integral accounts for the dimensions in the z, y and x direction respectively.
03

Putting It All Together

Once the limits of integration have been identified, the triple integral is \(\int_0^4 \int_0^{4-x} \int_0^{4-x-y} dz dy dx\), which represents the volume of the solid bounded by the plane and the first octant.

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In your own words, describe \(r\) -simple regions and \(\theta\) -simple regions.

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