The volume of the solid under the hyperbolic paraboloid [tex]z = 4 + x^2-y^2[/tex] and above the square r = [−1, 1] × [0, 2] is [tex]20\text{ cubic units}[/tex]
The question is asking us to find the volume bounded by the following surfaces
To do this, we have to evaluate the triple integral
[tex]\displaystyle \int\limit_0^2\,dx \int\limit_{-1}^1\,dy\int\limit_0^{4+x^2-y^2}\,dz[/tex]
First, we integrate with respect to [tex]z[/tex]
[tex]\displaystyle \int\limit_0^2\,dx \int\limit_{-1}^1\,dy\int\limit_0^{4+x^2-y^2}\,dz\\\\=\displaystyle \int\limit_0^2\,dx \int\limit_{-1}^1\,dy(4+x^2-y^2)[/tex]
Next, we integrate with respect to [tex]y[/tex]
[tex]\displaystyle \int\limit_0^2\,dx \int\limit_{-1}^1\,dy(4+x^2-y^2)\\\\=\displaystyle \int\limit_0^2\,dx \left[4y+x^2y-\dfrac{y^3}{3} \right]_{-1}^1\\\\=\displaystyle \int\limit_0^2\,dx \left[ \left(4(1)+x^2(1)-\dfrac{(1)^3}{3} \right)- \left(4(-1)+x^2(-1)-\dfrac{(-1)^3}{3} \right)\right]\\\\=\displaystyle \int\limit_0^2\,dx \left(\dfrac{22}{3}+2x^2 \right)\\\\[/tex]
Finally, we integrate with respect to [tex]x[/tex]
[tex]\displaystyle \int\limit_0^2\,dx \left(\dfrac{22}{3}+2x^2 \right)\\\\=\left[\dfrac{22}{3}x+\dfrac{2x^3}{3}\right]_0^2\\\\=\left(\dfrac{22}{3}(2)+\dfrac{2(2)^3}{3}\right) - \left(\dfrac{22}{3}(0)+\dfrac{2(0)^3}{3}\right)\\\\=20 \text{ cubic units}[/tex]
The volume of the solid is [tex]20\text{ cubic units}[/tex]
Learn more about volume of solids here https://brainly.com/question/6957600
