Question

Find the volume of the solid that lies under the hyperbolic paraboloid z=3y2−x2+6 and above the rectangle R=[−1,1]×[1,2].

Answer 1

Answer:

The volume of the solid is $\frac{76}{3} u^3$

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  $[-1,1] X [1,2]$ is taken in the plane XY. In this way the volume of the solid is:

$\int _1^2\left[\int _{-1}^1\:\left(3y^2-x^2+6\right)dx\right]dy = \int _1^2[12-\frac{2}{3}+6y^2]dy\right = \frac{76}{3} u^3$