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Chapter 13: Problem 24

Use a double integral to find the volume of the solid bounded by the graphs of the equations. $$ z=x, z=0, y=x, y=0, x=0, x=4 $$

Short Answer

Expert verified
The volume of the solid is 32 units.

Step by step solution

01

Identifying the bounds

First of all, identify the limits of integration from the given equations. Here, \(z=x\) and \(z=0\) show that \(z\) ranges from \(0\) to \(x\). \(y=x\) and \(y=0\) show that \(y\) ranges from \(0\) to \(x\). \(x=0, x=4\) indicate that \(x\) ranges from 0 to 4.
02

Setting up the integral

The volume of the solid can be computed by integrating the function that represents the upper surface minus the function that represents the lower surface in this case from z=0 to z=x. Since \(y\) is also independent of \(z\) but dependent on \(x\), we can integrate \(y\) from 0 to \(x\), and \(x\) from 0 to 4. The volume, \(V\), therefore is: \(V = \int_0^4\int_0^x\int_0^xdzdydx\).
03

Applying the integration

Perform the integration step by step. For any integral, start integrating from the innermost integral. In this case, integrating z over 0 to x is the first step: after the first integral, we get \(V = \int_0^4\int_0^x \frac{x^2}{2} dy dx\). Integrating the middle term, we get \(V = \int_0^4 \frac{x^3}{2}dx\). Integrating the outermost term we get: \(V = \left. \frac{x^4}{8}\right|_0^4 \).
04

Evaluating the integral

Evaluate the definite integral at the boundaries. \(V = \left. \frac{x^4}{8}\right|_0^4 \) becomes \(V = \frac{4^4}{8} = 32\). There was no need to subtract the lower limit \(x=0\) as it would have resulted to zero.

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Most popular questions from this chapter

Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{-2}^{2} \int_{0}^{4-y^{2}} d x d y $$

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