integrationmultivariable-calculus
Suppose R is the solid region bounded by the plane $z = 4x$, the surface $z = x^2$, and the planes $y = 0$ and $y = 3$. Write an iterated integral in the form below to find the volume of the solid R.
$$\iiint\limits_Rf(x,y,x)\mathrm{d}V=\int_A^B {\int_C^D {\int_E^F\mathrm{d}z} \mathrm{d}y} \mathrm{d}x$$
I need to find the limits, I found A and C which are zeros and I could not find the rest of the limits
To formalize my comments, I will put it in answer form. First off, your solution is correct. Just to add some clarification, the $xy$ plane is equivalently represented as $z=0$. The triple integral, $\int \int \int_0^{x^2-y^2} dzdydx$ reduces to a double integral that you described above. Next, we take the projection of the surface on the $xy$ plane, by finding the intersection of $z=x^2-y^2$ and $z=0$ which is simply $x^2=y^2$. Following that we draw the projection and deduce the following,
$$V=\int_{2}^{4} \int_{-x}^{x} \int_0^{x^2-y^2} dzdydx$$
The projection should look something like this (the red line is the ray that is drawn to determine the bounds)
Your bounds and integrand are correct, but here we can take a shortcut. The integral finds the volume under a rectangular section of the surface $z=\sin y$, which does not depend on $x$. Therefore we can just take the one-dimensional integral $$\int_0^{\pi/2}\sin y\ dy=1$$ and multiply it with the extent of the solid in the $x$-axis, which is also 1, to get the final answer of 1.
Best Answer
The region in xz plane is described by $0 \le x \le 4$ and $ x^2\le z\le 4x$
Thus the volume is found by $$V=\int_{0}^4 \int_{0}^{3} \int_{x^2}^{4x} dzdydx$$