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The volume is given by the triple integral,

[tex]\displaystyle\int_0^2\int_0^2\int_0^{4-x^2}\mathrm dz\,\mathrm dx\,\mathrm dy[/tex]

Easy enough to compute. Since the integrand is independent of [tex]y[/tex], you can compute that integral right away to get

[tex]2\displaystyle\int_0^2\int_0^{4-x^2}\mathrm dz\,\mathrm dx[/tex]

Next, with respect to [tex]z[/tex],

[tex]2\displaystyle\int_0^2z\bigg|_{z=0}^{z=4-x^2}\,\mathrm dx=\int_0^2(4-x^2)\,\mathrm dx[/tex]

which results in a volume of

[tex]2\displaystyle\left(4x-\frac13x^3\right)\bigg|_{x=0}^{x=2}=2\left(8-\frac83\right)=\dfrac{32}3[/tex]

We use triple integration to find volume of solid.

Volume of the solid in the first octant is 32/ 3

To solve this, it is important to set up the boundaries of double integral correctly.

First, notice that the cylinder we are given intersects the XY plane at the line given by the equation x = 2

Hence our boundaries of the integral with respect to x is the segment [0,2]. Since the other plane is y=2 . so,  boundaries of the integral with respect to y is the segment [0,2].

Now , integrate that

                          [tex]\int\limits^2_0 \int\limits^2_0 (4-x^{2} )dxdy\\\\\frac{16}{3} \int\limits^2_0 \, dy\\\\\frac{16}{3}*2\\\\=\frac{32}{3}[/tex]

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https://brainly.com/question/20284914