Evaluate the triple integral of (2x − 4yz) dV over the bounded region E where 2 ≤ x ≤ 3, 0 ≤ y ≤ 2x − 4 and 0 ≤ z ≤ 5x + 5y − 3. Enter an exact answer.

Step-by-step solution

1. 1

   Check that the description really is a solid. The limits are already nested in the order $dz\,dy\,dx$, so nothing has to be re-ordered — but they only describe a genuine region if each upper limit beats its lower one. On $2\le x\le 3$ the value $2x-4$ runs from $0$ to $2$, so the $y$-range is non-empty, and $5x+5y-3\ge 5(2)-3=7>0$, so the $z$-range is non-empty too. The integral is therefore

   $\int_{2}^{3}\!\int_{0}^{2x-4}\!\int_{0}^{5x+5y-3}(2x-4yz)\,dz\,dy\,dx.$

2. 2

   Integrate in $z$ first, because the integrand is only quadratic there. With $x$ and $y$ held fixed, an antiderivative of $2x-4yz$ is $2xz-2yz^{2}$. Writing $Z=5x+5y-3$ for the upper limit,

   $\int_{0}^{Z}(2x-4yz)\,dz = 2xZ-2yZ^{2}.$

   Keeping the limit packaged as $Z$ avoids expanding a cube by hand later.

3. 3

   Expand into a polynomial in $x$ and $y$. Since $Z^{2}=25x^{2}+25y^{2}+9+50xy-30x-30y$,

   $2xZ = 10x^{2}+10xy-6x,$
   $-2yZ^{2} = -50x^{2}y-100xy^{2}-50y^{3}+60xy+60y^{2}-18y.$

   Adding them and collecting the two $xy$ terms gives the inner integrand

   $10x^{2}+70xy-6x-50x^{2}y-100xy^{2}+60y^{2}-50y^{3}-18y.$

4. 4

   Integrate in $y$ from $0$ to $a=2x-4$. Term by term,

   $(10x^{2}-6x)a+35xa^{2}-25x^{2}a^{2}-\tfrac{100x}{3}a^{3}+20a^{3}-\tfrac{25}{2}a^{4}-9a^{2}.$

   Now substitute $a=2x-4$ and collect powers of $x$. This is the step where a sign or a binomial coefficient is easiest to lose, so expand $a^{2}=4x^{2}-16x+16$, $a^{3}=8x^{3}-48x^{2}+96x-64$ and $a^{4}=16x^{4}-128x^{3}+384x^{2}-512x+256$ explicitly. The result is

   $-\frac{1700}{3}x^{4}+3920x^{3}-10008x^{2}+\frac{33544}{3}x-4624.$

5. 5

   Integrate that quartic from $x=2$ to $x=3$. An antiderivative is

   $F(x)=-\frac{340}{3}x^{5}+980x^{4}-3336x^{3}+\frac{16772}{3}x^{2}-4624x,$

   and $F(3)-F(2)=-268$. Because every coefficient is rational, the answer is an exact integer with no rounding involved.

6. 6

   Confirm with an independent numerical quadrature. Evaluating the same iterated integral with a 60-point Gauss–Legendre rule in each variable returns $-268.0000000000$, matching the exact value. This check matters here: a common shortcut is to collapse step 4 mentally, and doing so gives the plausible-looking but wrong quartic $-\tfrac{800}{3}x^{4}+1840x^{3}-4700x^{2}+\tfrac{15968}{3}x-2048$, whose integral is $+416/3$.