Mathematics, 05.09.2019 00:10 magicalforlife
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].
Answers: 3
![[-1,1] X [1,2]](/tpl/images/0223/0674/dca55.png)
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](/tpl/images/0223/0674/e5c1a.png)
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Find the volume of the solid that lies under the hyperbolic paraboloid z=3y2−x2+6 and above the rect...
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