Calculate $$\iiint_V (15x+30z)\,dx\,dy\,dz$$ where $$V: \begin{cases}z = x^2+3y^2,\ \ z = 0 \\ y=x,\ \ y=0,\ \ x=1 \end{cases}$$
- I tried $x = r\cos\left(t\right), \ \ y = r\sin\left(t\right), \ \ z = h$, then obviously $t \in \left[0, \pi/4\right]$ and $h \in \left[0, r^{2}\left[\cos^{2}\left(t\right) + 3\sin^{2}\left(t\right)\right]\right]$.
- However, I cannot set the interval for $r$.
- The triangular shape formed by $x=1$ makes it difficult for me.
You have:
$x = r\cos\theta\\y=r\sin\theta\\z=h$
Substitute these into the equations for the boundary.
$h = r^2\cos^2 \theta + 3r^2\sin^2\theta\\ r\sin\theta = r\cos\theta\\ r\sin\theta = 0\\ r\cos\theta = 1$
What can we simplify?
$\sin\theta = \cos\theta\\ \theta= \frac {\pi}{4}$
$r\sin\theta = 0\\ \theta = 0,\text{ or } r = 0$
But you have all of these.
As for that boundary for $r.$
$r\cos\theta = 1\\ r = \sec\theta$
$\int_0^{\frac {\pi}{4}}\int_0^{\sec\theta}\int_0^{r^2(\cos^2\theta + 3\sin^2\theta)} r\ dz\ dr\ d\theta$