Volume integration

Question

Let

$$D = \left\{(x,y,z)\in\mathbb{R}^{3}\mid x\ge0,0\le y\le x, x^2+y^2\le {16}, 0\le z\le {5}\right\}.$$

I want to integrate

$$\displaystyle\iiint\limits_{D}\left({-4\,z+y^2+x^2}\right)\,\mathrm{d}V $$

We can see that $x^2+y^2=r^2$ so $r^2=16$. $r\to[0,16]$ and $\theta\to[0,2\pi]$ and $z \to[0,5]$

And then integration

$$\int_0^{2\pi}\int_0^5\int_0^{16}(-4z+y^2+x^2)\,dV= \int_0^{2\pi}\int_0^5\int_0^{16}(-4z+r^2)\,drdzd\theta=\frac{36160\pi}{3}$$

That is wrong answer and I don't know where I have done mistake.

Answer 1

You have some error in the bounds for $r$ and $\theta$, indeed in cylindrical coordinates we have

$$\iiint\limits_{D}\left({-4\,z+y^2+x^2}\right)\,\mathrm{d}V=$$

$$=\int_0^{\pi/4}d\theta\int_0^4rdr\int_0^5(-4z+r^2)dz$$

Answer 2

There are two errors.

• $r^2=16$ means $r=4$ not $16$
• $x\ge 0$ and $0\le y\le x$ means that $\theta$ varies between $0$ and $\pi/4$

Answer 3

You have an equation of a Circle in the xy-plane with radius 4. Your limits of integration with respect to radius lies only between 0 and 4 inclusive (i.e [0,4]).Clearly, x is positive and y lies between 0 and x inclusive (i.e [0,x]). Thus, your limits of Integration with respect to $\theta$ lies from $0$ to $\frac{\pi}{4}$When you integrate in cylindrical coordinates, remember to tag on an extra “r”. Your limits of integration with respect to the altitude of the cylinder is correct though. Your bounds therefore are; $$\int_0^5 \int_0^\frac{\pi}{4} \int_0^4 r(r^2 - 4z) dr d\theta dz$$