2
$\begingroup$

The volume of the solid body bounded by $x^2+y^2=4$ and the planes $y+z=4$ , $z=0$ should be calculated. The class notes say that this type of problem is solved using volume integral $\iiint \limits_G dV $.

Work so far:

**Edit (based on tom's inputs)

$ \iint\limits_R \big[\int \limits_0^{4-y}dz\big] dA = \int\limits_{-2}^2 \int\limits_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int\limits_0^{4-y} dz\,dy\,dx = 16\pi$

I need help with figuring out limits of integration on R, it is not clear from the examples given in class.

  • 0
    I solved it and got $16\pi$, thanks!2011-11-20

1 Answers 1

4

So tom is right, but if you attack the problem directly, then the integral is difficult.

Let $h_1, h_2$ be the min and max heights of the cylinder. ($2$ and $6$ resp.) Then $ h(x,y) = h_1 + \frac{1}{2}(h_2 - h_1)(1 + \frac{y}{2})$ $ h(r,\theta) = h_1 + \frac{1}{2}(h_2 - h_1)(1 + \frac{r\sin(\theta)}{R}) $ $ V = \int_0^{2\pi} \int_0^R r*h(r,\theta)\, dr\, d\theta = \frac{1}{2} \pi R^2 (h_1 + h_2) $