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Krzychau1

Answered question

2021-12-25

Find the volume of the solid that lies under the hyperbolic paraboloid

z = 3 y^{2} - x^{2} + 2

and above the rectangle

R = [- 1, 1] \times [1, 2]

Answer & Explanation

Maricela Alarcon

Beginner2021-12-26Added 28 answers

Volume of hyperbolic paraboloid

z = 3 y^{2} - x^{2} + 2

above the rectangle

[- 1, 1] \times [1, 2]

= \int_{- 1}^{1} \int_{1}^{2} (3 y^{2} - x^{2} + 2) d y d x

= \int_{- 1}^{1} (\int_{1}^{2} 3 y^{2} - x^{2} + 2 d y) d x

= \int_{- 1}^{1} {{[y^{3} - y x^{2} + 2 y]}_{1}^{2}} d x

= \int_{- 1}^{1} 9 - x^{2} d x

= {[9 x - \frac{x^{3}}{3}]}_{- 1}^{1} = 18 - \frac{2}{3} = \frac{52}{3}

Result:
Volume

= \frac{52}{3}

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