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I am trying to determine the bounds given a solid bounded by the plane $y+z=4$ and the cylinder $y=x^2$, and $xz$ and $yz$ planes in the first octant. I am not sure if I am on the right track, but to find my bounds for $y$, I set $z$ equal to zero and calculated $y=4$. This gives a region enclosed by a parabola with the upper y-bound at 4. So my outer integral (assuming type I region), I believe, is $x^2 \leq y \leq 4$. And since there is a vertical asymptote at $x=1$, I am assuming that my inner integral is bounded by $0 \leq x \leq 1$.

I know that I am going to need to integrate xz and yz as the functions, but I am not sure how to express these in terms of $z$, which is what I assume I am supposed to do since all problems I have encountered up until this point that are in 3D have done this. However, all of those problems didn't have z as a coefficient of either the x or y terms (e.g. they were in a form like $x + y + z$). Am I on the right track? And if so, how should I approach this?

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2 Answers 2

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Your first paragraph is almost spot on. I cannot make sense of your second paragraph. The mistake in the first paragraph is that you are making up the limit for $x$. The intersection between $y=x^2$ and $y=4$ (the intersection of the plane $y+z=4$ and the $x$-$y$ plane).

Since we have to stay in the first octant, the solid is bounded below by the $x$-$y$ axis.

The volume of your solid is $$ \int_0^2\int_{x^2}^4\,(4-y)\,dy\,dx. $$ Properly speaking, your integrand is $(4-y)-0$, as you integrate the difference between the "ceiling" and the "floor".

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I see where my mistake is now. You are right. In fact, I don't know why I thought a parabola was asymptotic because it clearly is not upon second look. –  Dylan Jun 12 '12 at 3:20
    
Also, I forgot the solid is also bounded by the xz and yz planes... Does this mean I have to do this again setting y to zero and x to zero, respectively? –  Dylan Jun 12 '12 at 3:43
    
Actually, I don't understand the reference to the xz and yz planes. The first octant is bounded by the xy, xz, and yz planes already, so there is no need for them to be mentioned in the question. –  Martin Argerami Jun 12 '12 at 14:32

It is useful to sketch, or plot the volume and domain of integration to get a better picture. What you suggested will work, the only bit you are missing is the function $z$ which is merely $$z=4-x$$ When you compute a double integral, you actually sum over a set of tall thin cylinders with bases mounted on the domain of integration and heights given by the values of $z$

enter image description here

It would be more convenient though to swap the order of integration and integrate first with respect to $x$ which, for every given $0\le y\le 4$ ranges from $0$ to $\sqrt{y}$. That is, you slice the parabola in stripes parallel to the $x$ axis.

Therefore,

$$V=\int_0^4 dy \int_0^{\sqrt{y}}\left(4-y\right)dx=\int_0^4\left(4y^\frac{1}{2}-y^\frac{3}{2}\right)dy=\left.\left(\frac{8}{3}y^\frac{3}{2}-\frac{2}{5}y^\frac{5}{2}\right)\right|_0^4=\frac{128}{15}$$ Now you can try to calculate using your original idea of the limits and verify, if you arrive at the same result. enter image description here

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Yes, this is the solution I got from integrating with the approach I was taking. Thanks. –  Dylan Jun 12 '12 at 3:18

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