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Chapter 13: Problem 55

Use a triple integral to compute the volume of the following regions. The wedge of the square column \(|x|+|y|=1\) created by the planes \(z=0\) and \(x+y+z=1\)

Short Answer

Expert verified
Based on the provided step by step solution, provide a short answer to find the volume of the region intersected by \(|x|+|y|=1\), \(z=0\), and \(x+y+z=1\): The volume of the region is found by first looking at the region in the first quadrant and finding the limits of integration. Then, the triple integral is set up and evaluated. Since the region is symmetrical, the volume of the first quadrant is multiplied by 4 to find the total volume. The volume of the given region is \(\frac{2}{3}\).

Step by step solution

01

Sketch the region

Before finding the limits of integration, it is helpful to sketch the region. The given region is a square column formed by the intersection of \(|x|+|y|=1\), \(z=0\), and \(x+y+z=1\). Notice that the plane\(x+y+z=1\) intersects the \(z = 0\) plane at the line \(x + y = 1\). And the square column has vertices at points \((1,0,0)\), \((-1,0,0)\), \((0,1,0)\), \((0,-1,0)\) , \((1,0,1)\), \((-1,0,1)\), \((0,1,1)\), \((0,-1,1)\). We can then find the four edges of the square column along the \(z = 0\) plane to understand the limits on x and y.
02

Find the limits of integration

Now we can find the limits of integration. The lower limit for z is given by the plane \(z=0\). The upper limit for z is given by the plane \(x+y+z=1 \Rightarrow z = 1-x-y\). Let's integrate z first. For the x and y limits, consider the square column in the xy-plane. The region formed by the intersection is as follows: \[ |x|+|y| = 1 \] Considering the first quadrant, where \(x \ge 0\) and \(y \ge 0\), we get \(x+y=1 \Rightarrow y = 1 - x\). The limits for x in the first quadrant are from 0 to 1 and for y are from 0 to \(1-x\). Since the region is symmetrical for all quadrants, we can multiply the volume of the first quadrant by 4 to obtain the total volume.
03

Set up the triple integral

Now that we have the limits of integration, it is time to set up the triple integral. The integrand is 1, since we are finding the volume. Total volume can be found using the following integral: \(V = 4\int\int\int dV\) For the first quadrant: \(V = 4\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}dzdydx\)
04

Evaluate the triple integral

Now we will evaluate the triple integral: \(V = 4\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y} 1 dzdydx\) First, integrate with respect to z: \(V = 4\int_{0}^{1}\int_{0}^{1-x}[z]|_{0}^{1-x-y} dydx\) \(V = 4\int_{0}^{1}\int_{0}^{1-x}(1-x-y) dydx\) Next, integrate with respect to y: \(V = 4\int_{0}^{1}[y(1-x)-\frac{1}{2}y^2]|_{0}^{1-x} dx\) \(V = 4\int_{0}^{1}[(1-x)^2-\frac{1}{2}(1-x)^2] dx\) \(V = 4\int_{0}^{1}\frac{1}{2}(1-x)^2 dx\) Finally, integrate with respect to x: \(V = 4[\frac{1}{6}x^3-x^2+x]|_{0}^{1}\) \(V = 4(\frac{1}{6}-1+1)\) \(V = \frac{4}{6}\) \(V = \frac{2}{3}\). Therefore, the volume of the given region is \(\frac{2}{3}\).

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Most popular questions from this chapter

Linear transformations Consider the linear transformation \(T\) in \(\mathbb{R}^{2}\) given by \(x=a u+b v, y=c u+d v,\) where \(a, b, c,\) and \(d\) are real numbers, with \(a d \neq b c\) a. Find the Jacobian of \(T\) b. Let \(S\) be the square in the \(u v\) -plane with vertices (0,0) \((1,0),(0,1),\) and \((1,1),\) and let \(R=T(S) .\) Show that \(\operatorname{area}(R)=|J(u, v)|\) c. Let \(\ell\) be the line segment joining the points \(P\) and \(Q\) in the uv- plane. Show that \(T(\ell)\) (the image of \(\ell\) under \(T\) ) is the line segment joining \(T(P)\) and \(T(Q)\) in the \(x y\) -plane. (Hint: Use vectors.) d. Show that if \(S\) is a parallelogram in the \(u v\) -plane and \(R=T(S),\) then \(\operatorname{area}(R)=|J(u, v)| \operatorname{area}(S) .\) (Hint: Without loss of generality, assume the vertices of \(S\) are \((0,0),(A, 0)\) \((B, C),\) and \((A+B, C),\) where \(A, B,\) and \(C\) are positive, and use vectors.)

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