Evaluate the triple integral of -5y with x from -1 to 1, y from 0 to sqrt(9 - x^2), and z from 0 to 9 - x^2 - y^2. Enter an exact answer.

Step-by-step solution

1. 1

   Integrate in z first, since the integrand has no z. With $-5y$ constant in $z$, the inner integral is just the integrand times the height of the column: $\int_0^{9-x^2-y^2}(-5y)\,dz=-5y\left(9-x^2-y^2\right)$.

2. 2

   Introduce a shorthand for the x-dependent radius. Set $a^2=9-x^2$, so the $y$-limits run from $0$ to $a$ and the integrand becomes $-5y(a^2-y^2)=-5\left(a^2y-y^3\right)$. Naming $a$ keeps the algebra readable and shows why the $y$-integral has a clean form.

3. 3

   Do the y-integral. $\int_0^{a}\left(a^2y-y^3\right)dy=\frac{a^2\cdot a^2}{2}-\frac{a^4}{4}=\frac{a^4}{4}$, so the double-inner result is $-\frac{5a^4}{4}=-\frac54\left(9-x^2\right)^2$.

4. 4

   Expand and use evenness in x. $\left(9-x^2\right)^2=81-18x^2+x^4$ is even, so $\int_{-1}^{1}=2\int_{0}^{1}$. The outer integral is $-\frac54\cdot 2\int_0^1\left(81-18x^2+x^4\right)dx$.

5. 5

   Evaluate the last integral. $\int_0^1\left(81-18x^2+x^4\right)dx=\left[81x-6x^3+\frac{x^5}{5}\right]_0^1=81-6+\frac15=\frac{376}{5}$.

6. 6

   Combine to get the exact value. $-\frac54\cdot 2\cdot\frac{376}{5}=-\frac52\cdot\frac{376}{5}=-188$.

7. 7

   Numerical check. Adaptive numerical integration of the original triple integral returns $-188.000000$ (estimated error $2\times10^{-12}$), matching the exact value.