find the volume of the solid that lies under the hyperbolic paraboloid ( z = 3y^{2}-x^{2}+2 ) and above the…

Explanation:

Step1: Set up the double - integral

The volume (V) of the solid under (z = f(x,y)=3y^{2}-x^{2}+2) and above the rectangle (R=[a,b]\times[c,d]=[-1,1]\times[1,2]) is given by the double - integral (V=\int_{c}^{d}\int_{a}^{b}f(x,y)dxdy). So, (V = \int_{1}^{2}\int_{-1}^{1}(3y^{2}-x^{2}+2)dxdy).

Step2: Integrate with respect to (x)

First, integrate (\int_{-1}^{1}(3y^{2}-x^{2}+2)dx). Using the integral rules (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)) and (\int kdx=kx + C) ((k) is a constant), we have: (\int_{-1}^{1}(3y^{2}-x^{2}+2)dx=\left[3y^{2}x-\frac{x^{3}}{3}+2x\right]_{x=-1}^{x = 1}) (=(3y^{2}(1)-\frac{1^{3}}{3}+2(1))-(3y^{2}(-1)-\frac{(-1)^{3}}{3}+2(-1))) (=(3y^{2}-\frac{1}{3}+2)-(-3y^{2}+\frac{1}{3}-2)) (=3y^{2}-\frac{1}{3}+2 + 3y^{2}-\frac{1}{3}+2) (=6y^{2}+\frac{-1 - 1}{3}+4) (=6y^{2}-\frac{2}{3}+4=6y^{2}+\frac{10}{3})

Step3: Integrate with respect to (y)

Now, integrate (\int_{1}^{2}(6y^{2}+\frac{10}{3})dy). Using (\int y^{n}dy=\frac{y^{n + 1}}{n + 1}+C(n\neq-1)) and (\int kdy=ky + C) ((k) is a constant): (\int_{1}^{2}(6y^{2}+\frac{10}{3})dy=\left[6\times\frac{y^{3}}{3}+\frac{10}{3}y\right]{1}^{2}) (=\left[2y^{3}+\frac{10}{3}y\right]{1}^{2}) (=(2(2)^{3}+\frac{10}{3}(2))-(2(1)^{3}+\frac{10}{3}(1))) (=(16+\frac{20}{3})-(2+\frac{10}{3})) (=16+\frac{20}{3}-2-\frac{10}{3}) (=14+\frac{20 - 10}{3}) (=14+\frac{10}{3}=\frac{42 + 10}{3}=\frac{52}{3})

Answer:

(\frac{52}{3})