Answer:
The volume of the solid is [tex]\frac{76}{3} u^3[/tex]
Step-by-step explanation:
The requested volume can be calculated through an iterated integral of the function that defines the solid. For such an iterated integral, the integration region defined by the rectangle [tex][-1,1] X [1,2][/tex] is taken in the plane XY. In this way the volume of the solid is:
[tex]\int _1^2\left[\int _{-1}^1\:\left(3y^2-x^2+6\right)dx\right]dy = \int _1^2[12-\frac{2}{3}+6y^2]dy\right = \frac{76}{3} u^3[/tex]
The volume of a shape is the amount of space in it.
The volume of the solid that lies under the hyperbolic paraboloid is 76/3
The given parameters are:
[tex]\mathbf{R=[-1,1] \times [1,2]}[/tex]
[tex]\mathbf{z =3y^2 - x^2 + 6}[/tex]
The volume of the solid is:
[tex]\mathbf{V = \int\limits^a_b {\int\limits^a_b z} \, dR }[/tex]
So, we have:
[tex]\mathbf{V = \int\limits^2_1 \int\limits^1_{-1} 3y^2 - x^2 + 6 \, dx \ dy}[/tex]
Integrate with respect to x
[tex]\mathbf{V = \int\limits^2_1 ( 3y^2x - \frac{x^3}{3} + 6x)|\limits^1_{-1} \ dy}[/tex]
Expand
[tex]\mathbf{V = \int\limits^2_1 ( 3y^2(1) - \frac{1^3}{3} + 6(1)) -( 3y^2(-1) - \frac{(-1)^3}{3} + 6(1-)) \ dy}[/tex]
[tex]\mathbf{V = \int\limits^2_1 ( 3y^2 - \frac{1}{3} + 6) -( -3y^2 + \frac{1}{3} -6) \ dy}[/tex]
[tex]\mathbf{V = \int\limits^2_1 ( 6y^2 - \frac{2}{3} + 12) \ dy}[/tex]
[tex]\mathbf{V = \int\limits^2_1 ( 6y^2 + \frac{-2+36}{3}) \ dy}[/tex]
[tex]\mathbf{V = \int\limits^2_1 ( 6y^2 + \frac{34}{3}) \ dy}[/tex]
Integrate with respect to y
[tex]\mathbf{V = (2y^3 + \frac{34}{3}y)|\limits^2_1 }[/tex]
Expand
[tex]\mathbf{V = (2\times 2^3 + \frac{34}{3}\times 2) - (2\times 1^3 + \frac{34}{3}\times 1)}[/tex]
[tex]\mathbf{V = (16 + \frac{68}{3}) - (2 + \frac{34}{3})}[/tex]
[tex]\mathbf{V = 14 + \frac{34}{3}}[/tex]
[tex]\mathbf{V = \frac{42+34}{3}}[/tex]
[tex]\mathbf{V = \frac{76}{3}}[/tex]
Hence, the volume of the solid is 76/3
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