I am studying for a test in calculus 3, and came across this question, the answer I get to is different from the one that did the test and got full score, but I can't understand what have I done wrong.
Let $E = \{(x,y,z,w) ; 0 \le w \le 1- (x^2 + y^2 +z^2)^3\}$
calculate the volume of E.
my attempt:
$\forall ~w \in [0,1]: Let~ E_w = \{(x,y,z,w) ;(x^2 + y^2 +z^2)^3 \le 1- w\} = \{(x,y,z,w) ;x^2 + y^2 +z^2 \le (1- w)^{\frac13}\}$
then $V(E_w) = V((1- w)^{\frac16} B_3(0,1))$ where $B_3(0,1)$ is a ball of radius 1 in $\mathbb{R}^3$ so $V(E_w) = \frac{4 \pi} {3} (1- w)^{\frac16}$
and $V(E) = \displaystyle \int\limits_0^1 V(E_w)\;dw = \frac{4 \pi} {3} \int\limits_0^1 (1- w)^{\frac16} =\frac{4 \pi} {3} \frac{6}{7} (1-w)^{\frac76}|_0^1 = \frac{8 \pi}{7}$
the answer in the test is $\frac{8 \pi}{9}$
where did I go wrong? or is my answer correct and the test guy just got lucky? thanks!
$0 \le x^2 + y^2 +z^2 \le (1- w)^{1/3}$
$\implies 0 \leq \rho \leq (1- w)^{1/6}$
$V(E_w) = \dfrac{4\pi}{3} \rho^3 = \dfrac{4\pi}{3} \sqrt{1-w}$
So, $ V(E) = \displaystyle \dfrac{4\pi}{3} \int_0^1 \sqrt{1-w} \ dw = \dfrac{8\pi}{9}$