Expert Answer to "Evaluating triple integrals: Each of the triple integrals..."

Brennan Flores

Answered question

2021-10-20

Evaluating triple integrals: Each of the triple integrals that follow represents the volume of a solid. Sketch the solid and evaluate the integral.

\int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} \int_{0}^{y} d z d y d x

Raheem Donnelly

Skilled2021-10-21Added 75 answers

Given:

\int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} \int_{0}^{y} d z d y d x

Now,

\int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} \int_{0}^{y} d z d y d x = \int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} {[z]}_{0}^{y} d y d x

= \int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} (y) d y d x

= \int_{- 2}^{2} {[\frac{y^{2}}{2}]}_{0}^{\sqrt{4 - x^{2}}} d x

= \int_{- 2}^{2} [\frac{{(\sqrt{4 - x^{2}})}^{2}}{2} - 0] d x

= \int_{- 2}^{2} [4 - x^{2}] d x

= {[4 x - \frac{x^{3}}{3}]}_{- 2}^{2}

= (4 (2) - \frac{{(2)}^{3}}{3}) - (4 (- 2) - \frac{{(- 2)}^{3}}{3})

(8 - \frac{8}{3}) - (- 8 + \frac{8}{3})

= 8 - \frac{8}{3} + 8 - \frac{8}{3}

= \frac{32}{3}

Hence,

\int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} \int_{0}^{y} d z d y d x = \frac{32}{3}

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