We are told to evaluate the triple integral:
$$\iiint_E z dV$$
where $E$ is bounded by $x=4y^2+4z^2$ and $x=4$.
My attempt:
First I noticed that this represents a paraboloid on the x axis so I thought to use cylindrical coordinates (however as the paraboloid was centered around x I wasnt sure whether to let $z=r\cos\theta$, $y=r\sin\theta$ or the other way around?) $$z=r\cos\theta$$ $$y=r\sin\theta$$ $$0<r<\frac{\sqrt x}{2}$$ $$0<\theta<2\pi$$ $$0<x<4$$ and our integral becomes $$\int_0^4\int_0^{2\pi}\int_0^{\frac{\sqrt x}{2}} r\cos\theta r dr d\theta dx=0$$
However my textbook James Stewart Calculus gives me $\frac{16\pi}{3}$.
Where have I gone wrong?
your answer for the integral $$\iiint_E z dV = 0$$ is correct since it is followed by the symmetry of $z$ in relation to the axis $x$.
instead, if you evaluate the integral: $$\iiint_E x dV$$ you will get: \begin{align*} \int_0^4\int_0^{2\pi}\int_0^{\frac{\sqrt x}{2}} x r \,dr \,d\theta \,dx &= \int_0^4\int_0^{2\pi}x^2\cdot\left(\frac{1}{8}\right)\,d\theta\,dx \\&=\int_0^4x^2\cdot\left(\frac{1}{8}\right)\cdot(2\pi)\, dx \\&=\frac{\pi}{4} \int_0^4x^2\,dx = \frac{16\pi}{3} \end{align*}