Mathematics: Find the volume of the solid that lies under the hyperbolic paraboloid z = 4 + x 2 − y 2 and and above the square r = [−1, 1] × [0, 2].

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Find the volume of the solid that lies under the

Ответ:

zakariaaap1

12.03.2022

Mathematics

The volume of the solid under the hyperbolic paraboloid z = 4 + x^2-y^2 and above the square r = [−1, 1] × [0, 2] is $20\text{ cubic units}$

The question is asking us to find the volume bounded by the following surfaces

The plane

and the surface z = 4 + x^2 - y^2

The planes x=-1

and

The planes y=0

and

To do this, we have to evaluate the triple integral

$\displaystyle \int\limit_0^2\,dx \int\limit_{-1}^1\,dy\int\limit_0^{4+x^2-y^2}\,dz$

Evaluating the triple integral

First, we integrate with respect to

$\displaystyle \int\limit_0^2\,dx \int\limit_{-1}^1\,dy\int\limit_0^{4+x^2-y^2}\,dz\\\\=\displaystyle \int\limit_0^2\,dx \int\limit_{-1}^1\,dy(4+x^2-y^2)$

Next, we integrate with respect to

$\displaystyle \int\limit_0^2\,dx \int\limit_{-1}^1\,dy(4+x^2-y^2)\\\\=\displaystyle \int\limit_0^2\,dx \left[4y+x^2y-\dfrac{y^3}{3} \right]_{-1}^1\\\\=\displaystyle \int\limit_0^2\,dx \left[ \left(4(1)+x^2(1)-\dfrac{(1)^3}{3} \right)- \left(4(-1)+x^2(-1)-\dfrac{(-1)^3}{3} \right)\right]\\\\=\displaystyle \int\limit_0^2\,dx \left(\dfrac{22}{3}+2x^2 \right)\\\\$

Finally, we integrate with respect to

$\displaystyle \int\limit_0^2\,dx \left(\dfrac{22}{3}+2x^2 \right)\\\\=\left[\dfrac{22}{3}x+\dfrac{2x^3}{3}\right]_0^2\\\\=\left(\dfrac{22}{3}(2)+\dfrac{2(2)^3}{3}\right) - \left(\dfrac{22}{3}(0)+\dfrac{2(0)^3}{3}\right)\\\\=20 \text{ cubic units}$