An exam type question regarding volume integrals, I know a bit how to solve them but some things are still unclear. especially because of the way this question was written
Let $K$ be the body in $\mathbb R^3$ be given by $$K= \{(x, y, z)| x \ge y^2, x-y\le 2 , 0 \le z \le x\}.$$ Find the volume of $K$.
My approach:
What I have tried is to first set-up the three integral signs but that's where it already goes wrong. You can see that you can integrate $z$ from $0$ to $x$ and with some shifting around you can see that you can integrate $x$ from $y^2$ to $2+y$. However the answers then say that you can also integrate $y$ from $-1$ to $2$. ( Why do they do this.
so you would get.
$$\int_{-1}^2\int_{y^2}^{y+2}\int_0^x \,dzdxdy=36/5$$
But this $36/5$ is just a guess. So my two main questions are, where does the $-1$,and $2$ come from and how did I get $36/5$?
The first thing I notice is that the first two conditions on K, $x\ge y^2$ and $x- y\le 2$ do not involve z so can be graphed in two dimensions. $x= y^2$ is the parabola with vertex (0, 0) opening to the right, x-axis as axis of symmetry. $x- y= 2$ is the line through (0, 2) and (2, 0). The region of integration is the region inside the parabola to the left of that line. $x= y^2$ and $x- y= 2$ intersect where $x- y= y^2- y= 2$. Completing the square, $y^2- y+ \frac{1}{4}= \frac{9}{4}$. $(y- 1/2)^2= 9/4$. $y- 1/2= \pm 3/2$, y= -1 and 2.
So we can take y from -1 to 2 and, for each y, x goes from $y^2$ to $y+ 2$.
Then we have the third condition, $0\le z\le x$ so for each x, y, z goes from 0 to x.
The integral is $\int_{y=-1}^2\int_{x= y^2}^{y+2}\int_{z= 0}^x dzdxdy= \int_{y=-1}^2\int_{x= y^2}^{y+2}x dxdy= \frac{1}{2}\int_{y=-1}^2 (y+2)^2-(y^2)^2 dy= \frac{1}{2}\int_{-1}^2 (y^2+ 4y+ 4- y^4) dy= \frac{1}{2}\left[\frac{1}{3}y^3+ 2y^2+ 4y- \frac{1}{5}y^5\right]_{-1}^2$.