Use a double integral to find the solid bounded by z = 1 - y^2 and z = y^2 -1 for 0 < x < 2

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where is the given series? question lacks data
In place of previous question I posted another one
answer 16/3
Dont know
i am telling that the answer is 16/3. the limits of integration are -1 to 1. please see, I modified.

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2014-10-15T15:37:09+05:30

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In the y-z plane, the curve z = 1 - y² and the curve z = y² - 1 form a bounded area around y axis. It is symmetric wrt y axis. The shape of curve is parabolic.

In the y-z plane the intersection of z = 1 - y
² and z = y² - 1 gives the limits of integration and define the bounded area.
 
            z = 1 - y
² = y² - 1       =>    2 y² = 2      =>  y = +1 or  -1

So the cross section area is bounded between  y = -1 and y = +1.

In the x-axis direction the solid has a uniform area.  So the solid is a cylinder with the length in the x direction with cross section in the y-z plane.

Volume = cross sectional area * length

Volume = \int\limits^2_0 {Cross-section-Area} \, dx = \int\limits^2_0 ({ \int\limits^a_b {(z_2-z_1)} \, dy }) \, dx \\ \\\int\limits^2_0 ({ \int\limits^1_{-1} {(1-y^2-(y^2-1))} \, dy }) \, dx \\ \\=2\int\limits^2_0 { \int\limits^1_{-1} {(1-y^2)} \, dy } \, dx \\ \\2 \int\limits^2_0 {[y-y^3/3]_{-1}^1} \, dx =2 \int\limits^2_0 {[1-1/3-(-1+1/3)]} \, dx\\ \\=2 *\frac{4}{3} \int\limits^2_0 {1} \, dx\\ \\=2*\frac{4}{3}*[x]_0^2=2*\frac{4}{3}*2 = 16/3\\

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