Solve for a in the triple integral r3-a-y² (4-x-y²14dz dx dy15Jo

the given expression is: ∫01∫03-a-y2∫a4-x-y2dz dx dy=1415 solve the right hand side of the given…

Answered: r3-a-y² (4-x-y² 14 dz dx dy 15 Jo

Algebra: Structure And Method, Book 1

(REV)00th Edition

ISBN:9780395977224

Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Chapter6: Fractions

Section6.1: Simplifying Fractions

Problem 29WE

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Question

Solve for a in the triple integral

Expert Solution

Step 1

the given expression is:

$\int_{0}^{1} \int_{0}^{3 - a - y^{2}} \int_{a}^{4 - x - y^{2}} d z d x d y = \frac{14}{15}$

solve the right hand side of the given triple integral:

L.H.S:

$\int_{0}^{1} \int_{0}^{3 - a - y^{2}} \int_{a}^{4 - x - y^{2}} d z d x d y = \int_{0}^{1} \int_{0}^{3 - a - y^{2}} {(z)}_{a}^{4 - x - y^{2}} d x d y = \int_{0}^{1} \int_{0}^{3 - a - y^{2}} (4 - x - y^{2} - a) d x d y = \int_{0}^{1} {(4 x - \frac{x^{2}}{2} - x y^{2} - a x)}_{0}^{3 - a - y^{2}} d y = \int_{0}^{1} {((4 - a - y^{2}) x - {\frac{x}{2}}^{2})}_{0}^{3 - a - y^{2}} d y = \int_{0}^{1} ((4 - a - y^{2}) (3 - a - y^{2}) - \frac{{(3 - a - y^{2})}^{2}}{2} - 0 - 0) d y = \int_{0}^{1} ((12 - 4 a - 4 y^{2} - 3 a + a^{2} + a y^{2} - 3 y^{2} + a y^{2} + y^{4}) - \frac{(9 + a^{2} + y^{4} - 6 a + 2 a y^{2} - 6 y^{2})}{2}) d y = \int_{0}^{1} (12 - 7 a + a^{2} - 7 y^{2} + 2 a y^{2} + y^{4} - (\frac{9 - 6 a + a^{2} + 2 a y^{2} - 6 y^{2} + y^{4}}{2})) d y = \int_{0}^{1} \frac{24 - 14 a + 2 a^{2} - 14 y^{2} + 4 a y^{2} + 2 y^{4} - 9 + 6 a - a^{2} - 2 a y^{2} + 6 y^{2} - y^{4}}{2} d y = \int_{0}^{1} \frac{15 - 8 a + a^{2} - 8 y^{2} + 2 a y^{2} + y^{4}}{2} d y = {(\frac{(15 - 8 a + a^{2}) y}{2} + \frac{(2 a - 8) y^{3}}{6} + \frac{y^{5}}{10})}_{0}^{1} = (\frac{15 - 8 a + a^{2}}{2} (1 - 0) + \frac{(2 a - 8) (1^{3} - 0^{3})}{6} + (\frac{1^{5} - 0^{5}}{10})) = \frac{15 - 8 a + a^{2}}{2} + (\frac{2 a - 8}{6}) + (\frac{1}{10}) = \frac{225 - 120 a + 15 a^{2} + 10 a - 40 + 3}{30} = \frac{15 a^{2} - 110 a + 188}{30}$

Step 2

as given $\int_{0}^{1} \int_{0}^{3 - a - y^{2}} \int_{a}^{4 - x - y^{2}} d z d x d y = \frac{14}{15}$

therefore, R.H.S= $\frac{14}{15}$

equate L.H.S=R.H.S

therefore,

$\begin{array}{rcl} \frac{15 a^{2} - 110 a + 188}{30} & = & \frac{14}{15} \\ 15 a^{2} - 110 a + 188 & = & 28 \\ 15 a^{2} - 110 + 160 & = & 0 \\ 3 a^{2} - 22 + 32 & = & 0 \\ 3 a^{2} - 6 a - 16 a + 32 & = & 0 \\ 3 a (a - 2) - 16 (a - 2) & = & 0 \\ (3 a - 16) (a - 2) & = & 0 \\ 3 a - 16 & = & 0 o r a - 2 = 0 \\ a & = & \frac{16}{3} o r a = 2 \end{array}$

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