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irina1246 [14]
3 years ago
14

Use the general slicing method to find the volume of the following solids. The solid whose base is the region bounded by the cur

ves y=x² and y=2−x², and whose cross sections through the solid perpendicular to the x-axis are squares

Mathematics
1 answer:
Y_Kistochka [10]3 years ago
5 0

Answer:

The volume is V=\frac{64}{15}

Step-by-step explanation:

The General Slicing Method is given by

<em>Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is</em>

V=\int\limits^b_a {A(x)} \, dx

Because a typical cross section perpendicular to the x-axis is a square disk (according with the graph below), the area of a cross section is

The key observation is that the width is the distance between the upper bounding curve y = 2 - x^2 and the lower bounding curve y = x^2

The width of each square is given by

w=(2-x^2)-x^2=2-2x^2

This means that the area of the square cross section at the point x is

A(x)=(2-2x^2)^2

The intersection points of the two bounding curves satisfy 2 - x^2=x^2, which has solutions x = ±1.

2-x^2=x^2\\-2x^2=-2\\\frac{-2x^2}{-2}=\frac{-2}{-2}\\x^2=1\\\\x=\sqrt{1},\:x=-\sqrt{1}

Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is

V=\int\limits^{1}_{-1} {(2-2x^2)^2} \, dx\\\\V=\int _{-1}^14-8x^2+4x^4dx\\\\V=\int _{-1}^14dx-\int _{-1}^18x^2dx+\int _{-1}^14x^4dx\\\\V=\left[4x\right]^1_{-1}-8\left[\frac{x^3}{3}\right]^1_{-1}+4\left[\frac{x^5}{5}\right]^1_{-1}\\\\V=8-\frac{16}{3}+\frac{8}{5}\\\\V=\frac{64}{15}

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We know that
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