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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.

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should that be 4-x instead of 4-y for the upper limit of the leftmost integral? – tom Nov 20 '11 at 13:52
Just as a general note, to define the region R it's generally easiest to provide a 'rough' outline for one coordinate, and then be more 'specific' with the other. For instance, in your situation, you might say $-2<x<2$, which is a very rough outline of the region, and then use the y coordinate to be more specific, maybe something like $-\sqrt{4-x^2}<y<\sqrt{4-x^2}$ – tom Nov 20 '11 at 13:56
Thanks, corrected to y+z = 4 in assignment text – 1osmi Nov 20 '11 at 13:58
I solved it and got $16\pi$, thanks! – 1osmi Nov 20 '11 at 19:18
up vote 4 down vote accepted

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 $$ h(x,y) = h_1 + \frac{1}{2}(h_2 - h_1)(1 + \frac{y}{2})$$ $$ h(r,\theta) = h_1 + \frac{1}{2}(h_2 - h_1)(1 + \frac{r\sin(\theta)}{R}) $$ $$ V = \int_0^{2\pi} \int_0^R r*h(r,\theta)\, dr\, d\theta = \frac{1}{2} \pi R^2 (h_1 + h_2) $$

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