1. ## Multiple integration

Solve for a

$\int_{0}^{1}\int_{0}^{4-a-x^2}\int_{0}^{4-x^2-y}dzdydx=\frac{4}{7}$

I started it like this:

The integration of the dz to get

$\int_{0}^{1}\int_{0}^{4-a-x^2}[4-x^2-y]dydx$

Then I integrated with respect to dy to get:

$\int_{0}^{1}[4y-x^2y-\frac{y^2}{2}]_{0}^{4-a-x^2}dx$

Then adding the limits and collecting the terms together I get:

$\int_{0}^{1}8-4x^2+a-\frac{a^2}{2}+x^4dx$

Then integrating it by dx

$[8x-\frac{4}{3}x^3+2ax-\frac{a^2x}{2}+\frac{x^5}{5}]_{0}^{1}$

$8-\frac{4}{3}+2a-\frac{a^2}{2}+\frac{1}{5}$

Putting the terms togther and making equal $\frac{4}{7}$

$105a^2-420a-481=0$

I think I have gone wrong somewhere but not sure where.

2. Originally Posted by LL_5
Solve for a

$\int_{0}^{1}\int_{0}^{4-a-x^2}\int_{0}^{4-x^2-y}dzdydx=\frac{4}{7}$

I started it like this:

The integration of the dz to get

$\int_{0}^{1}\int_{0}^{4-a-x^2}[4-x^2-y]dydx$

Then I integrated with respect to dy to get:

$\int_{0}^{1}[4y-x^2y-\frac{y^2}{2}]_{0}^{4-a-x^2}dx$

Then adding the limits and collecting the terms together I get:

$\int_{0}^{1}8-4x^2+a-\frac{a^2}{2}+x^4dx$

Then integrating it by dx

$[8x-\frac{4}{3}x^3+2ax-\frac{a^2x}{2}+\frac{x^5}{5}]_{0}^{1}$

$8-\frac{4}{3}+2a-\frac{a^2}{2}+\frac{1}{5}$

Putting the terms togther and making equal $\frac{4}{7}$

$105a^2-420a-481=0$

I think I have gone wrong somewhere but not sure where.
$[4y-x^2y-\frac{y^2}{2}]_{0}^{4-a-x^2}$

$= 4(4-a-x^2) - x^2(4-a-x^2) - \frac{(4-a-x^2)^2}{2}$

$= 16 - 4a -4x^2 - 4x^2+x^2a+x^4 - \frac{4(4-a-x^2) - a(4-a-x^2)-x^2(4-a-x^2)}{2}$

$= 16 - 4a -4x^2 - 4x^2+x^2a+x^4 + \frac{-16 +4a +4x^2+ 4a-a^2-x^2a+4x^2 - ax^2 -x^4}{2}$

$= 16 - 4a -4x^2 - 4x^2+x^2a+x^4 + -8 +2a +2x^2+ 2a-\frac{a^2}{2}-\frac{x^2a}{2}+2x^2 - \frac{ax^2}{2} -\frac{x^4}{2}$

$= 8-4x^2-\frac{a^2}{2} +\frac{x^4}{2}$

3. So what 's the value a?

Cos I still can't get it.

And I think I may be wrong in the way I did this question