$$\iint_D (6x+2y) \, \mathrm d x \,\mathrm d y$$

where $D$ is the convex hull of $4$ given points,

$$D = \mbox{conv} \left\{ (0,0),(-2,6),(3,2),(1,8) \right\}$$

This is a parallelogram with "unit vectors" $(-2,6)$, $(3,2)$.

I wanted to give a try to solve the following problem with algebra instead of calculus. So, I thought about calculating the area of it and piling it up to get its volume. I got the cosine between both vectors with the dot product formula, which is $$\dfrac{6}{\sqrt{13\cdot40}}$$ and the sine with the Pythagoras identity which is $$\sqrt{\dfrac{484}{13\cdot40}}$$ I remember that $|a|\cdot |b|\cdot \sin\alpha$ gives the height, and $|a|\cdot |b|\cdot \cos\alpha$ gives the area. So I figured that maybe this would solve the integral problem for the area?

$$\dfrac{6}{\sqrt{13\cdot40}} \cdot \sqrt{13} \cdot \sqrt{40}\sqrt{\dfrac{484}{{13\cdot40}}}\cdot \sqrt{13} \cdot \sqrt{40}$$

That gives me $$6 \cdot 22 = 132$$ That's wrong but the right result is $$11 \cdot 22 = 242$$ so there might be something in that?

  • $\begingroup$ You can compute an integral with the aid of areas if your integrand is 1 or a constant. In your case you have an integrand which contains other variables $x,y$... In this case you should decompose your domain and transform the double integral into two successive one dimensional integrals. Or change variables in order to transform your domain into a square (not sure that's easier to do...) $\endgroup$ Jul 6, 2018 at 9:37
  • 2
    $\begingroup$ May be the solid volume of which you want to get is not a parallelipiped. Its top is not parallel to XOY $\endgroup$ Jul 6, 2018 at 9:40
  • $\begingroup$ @BeniBogosel: would it work if I did a change of base from $$\begin{pmatrix} \begin{align} 3 & -2 \\ 2 && 6\\ \end{align} \end{pmatrix}$$ Sorry, I can't separate the 2 and the 6 properly! $\endgroup$
    – Dovendyr
    Jul 6, 2018 at 10:03
  • $\begingroup$ Maybe if I made a cube of the integral then I would be able to calculate it's volume then? $\endgroup$
    – Dovendyr
    Jul 6, 2018 at 10:12

2 Answers 2


The area of the domain is given by the cross product of two sides,


Then the integral must be the area times the average value of the integrand, which by linearity is also the value of the function at the centroid, $\left(\dfrac12,4\right)$, hence


Your answer cannot be right, because you make no use of the coefficients $6$ and $2$ in the integrand. What you are evaluating is


which has no justification.

Another solution is by noting that the volume is a truncated prism with bases $(0,0,0),$ $(-2,6,0),$ $(3,2,0),(1,8,0)$ and $(0,0,0),(-2,6,0),(3,2,22),(1,8,22)$, which is the half of a non-truncated prism of height $22$. Hence


  • $\begingroup$ Thanks @YvesDaoust, I really love your answer! How do I find the centroid? $\endgroup$
    – Dovendyr
    Jul 6, 2018 at 10:46
  • $\begingroup$ @Dovendyr: by symmetry, the centroid of the parallelogram is the centroid of the four vertices, i.e. the arithmetic mean. $\endgroup$
    – user65203
    Jul 6, 2018 at 10:46
  • $\begingroup$ I like your second take on it as well, but can you send some graphic representation?? $\endgroup$
    – Dovendyr
    Jul 6, 2018 at 10:47
  • $\begingroup$ @Dovendyr: sorry, no, your task. $\endgroup$
    – user65203
    Jul 6, 2018 at 10:47
  • $\begingroup$ I got an empty box. Even though I follow the instructions! Graphics3D[ Prism[ {0, 0, 0}, {-2, 6, 0}, {3, 2, 0}, {1, 8, 0}, {0, 0, 0}, {-2, 6, 0}, {3, 2, 22}, {1, 8, 22}]] $\endgroup$
    – Dovendyr
    Jul 6, 2018 at 10:56


$$\begin{bmatrix} x\\ y\end{bmatrix} = \begin{bmatrix} 3 & -2\\ 2 & 6\end{bmatrix} \begin{bmatrix} u\\ v\end{bmatrix}$$

where the determinant of the matrix is $22$. Since

$$6x + 2y = \begin{bmatrix} 6\\ 2\end{bmatrix}^\top \begin{bmatrix} x\\ y\end{bmatrix} = \begin{bmatrix} 6\\ 2\end{bmatrix}^\top \begin{bmatrix} 3 & -2\\ 2 & 6\end{bmatrix} \begin{bmatrix} u\\ v\end{bmatrix} = \begin{bmatrix} 22\\ 0\end{bmatrix}^\top \begin{bmatrix} u\\ v\end{bmatrix} = 22 u$$

we have

$$\iint_D (6x+2y) \, \mathrm d x \,\mathrm d y = \iint_{[0,1]^2 } (22 u) (22 \, \mathrm d u \,\mathrm d v) = 22^2 \int_0^1 u \, \mathrm d u = \frac{22^2}{2} = 22 \cdot 11$$

  • 1
    $\begingroup$ This is amazing! So neat! I have to give it some thought to understand how and why the determinant comes into play. I'll be back! $\endgroup$
    – Dovendyr
    Jul 6, 2018 at 13:15
  • $\begingroup$ Note that the gradient of $6x+2y$ is orthogonal to the 2nd column of the matrix. The integrand was not chosen randomly! $\endgroup$ Jul 6, 2018 at 13:35
  • $\begingroup$ I have been thinking a bit about that, but why is that : $$\begin{bmatrix} u\\ v\end{bmatrix} = \begin{bmatrix} 3 & -2\\ 2 & 6\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix}$$? Isn't it the vectors $u$ and $v$ that have coordinates $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ 6 \end{pmatrix}$ respectively? And in addition, why is that $6x+2y=22u$? Where is that coming from? $\endgroup$
    – Dovendyr
    Jul 7, 2018 at 9:14
  • 1
    $\begingroup$ You have a parallelogram in the $(x,y)$ plane. It is a pain to integrate a function over this parallelogram. To make your life easier, consider a unit square $[0,1]^2$ in the $(u,v)$ plane. Via the linear transformation $$\begin{bmatrix} x\\ y\end{bmatrix} = \begin{bmatrix} 3 & -2\\ 2 & 6\end{bmatrix} \begin{bmatrix} u\\ v\end{bmatrix}$$ the $4$ vertices of the unit square in $(u,v)$ are mapped to the $4$ vertices of the parallelogram. Fortunately, the integrand in terms of $u$ and $v$ depends only on $u$. Thus, perform the integration over the unit square of the $(u,v)$ plane. $\endgroup$ Jul 7, 2018 at 12:46
  • 1
    $\begingroup$ Note that the unit square in the $(u,v)$ plane is ($22$ times) smaller than the parallelogram in the $(x,y)$ plane. Thus, to compensate for the shrinking, multiply the integral over $u$ and $v$ by the determinant of the matrix (which is $22$). $\endgroup$ Jul 7, 2018 at 13:00

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