/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Suppose a wedge of cheese fills ... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose a wedge of cheese fills the region in the first octant bounded by the planes \(y=z, y=4,\) and \(x=4\) You could divide the wedge into two pieces of equal volume by slicing the wedge with the plane \(x=2 .\) Instead find \(a\) with \(0

Short Answer

Expert verified
Based on the presented solution, when the wedge is sliced by the plane y = a, which divides it into two pieces of equal volume, the value of a is √8.

Step by step solution

01

Calculate the Total Volume of the Wedge

To calculate the total volume of the wedge, we can use a triple integral over the region bounded by the planes. We will use the following integral: $$V_{total} = \iiint_{D} dV = \int_{0}^{4} \int_{0}^{4} \int_{0}^{y} dx\, dy\, dz$$ where \(D\) is the region described by the planes. Let's evaluate this integral:
02

Integrate with respect to x

$$V_{total} = \int_{0}^{4} \int_{0}^{4} \int_{0}^{y} 1 \,dx\, dy\, dz = \int_{0}^{4} \int_{0}^{4} [x]_{0}^{y} dy\, dz$$
03

Integrate with respect to y

$$V_{total} = \int_{0}^{4} \int_{0}^{4} y \,dy\, dz = \int_{0}^{4} \left[\frac{1}{2}y^2\right]_{0}^{4} dz$$
04

Integrate with respect to z

$$V_{total} = \int_{0}^{4} \left[8\right] dz = 8\left[z\right]_{0}^{4}$$ So, the total volume of the wedge is $$V_{total} = 8(4) = 32$$
05

Calculate the Volume of One Piece of the Wedge

The volume of one piece of the wedge is determined when the plane \(y=a\) divides it. We can write the integral for this piece in the following manner: $$V_{left} = \iiint_{D} dV = \int_{0}^{4} \int_{0}^{a} \int_{0}^{y} dx\, dy\, dz$$
06

Integrate with respect to x

$$V_{left} = \int_{0}^{4} \int_{0}^{a} \int_{0}^{y} 1 \,dx\, dy\, dz = \int_{0}^{4} \int_{0}^{a} [x]_{0}^{y} dy\, dz$$
07

Integrate with respect to y

$$V_{left} = \int_{0}^{4} \int_{0}^{a} y \,dy\, dz = \int_{0}^{4} \left[\frac{1}{2}y^2\right]_{0}^{a} dz$$
08

Integrate with respect to z

$$V_{left} = \int_{0}^{4} \left[\frac{1}{2}a^2\right] dz = \frac{1}{2}a^2[z]_{0}^{4}$$ So the volume of one piece of the wedge is $$V_{left} = 2a^2$$
09

Find the value of \(a\)

Now we must find the value of \(a\) such that the volume of the left piece of the wedge is half of the total volume. $$V_{left} = \frac{1}{2}V_{total}$$ By substituting the values for \(V_{left}\) and \(V_{total}\), $$2a^2 = \frac{1}{2}(32)$$ Now, solve for a: $$2a^2 = 16 \Rightarrow a^2 = 8 \Rightarrow a = \sqrt{8}$$ Therefore, the value of \(a\) that divides the wedge into two pieces of equal volume is: $$a = \sqrt{8}$$.

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