Evaluate the triple integral of 1 dz dy dx where z runs from 0.9621 - sqrt(0.9649^2 - (x - 0.4167)^2) to 0.9621 + sqrt(0.9649^2 - (x - 0.4167)^2), y runs from 1.7476 - sqrt(1.7476^2 - x^2) to 0.6773, and x runs from 0.34356 to 1.3815.

Step-by-step solution

1. 1

   Read the integral as a volume. The integrand is the constant $1$, so the value of the triple integral is simply the volume of the region described by the three pairs of limits. That reframing is useful because it tells you the answer must come out positive, and it gives a rough size to sanity-check against later.

2. 2

   Do the $z$-integral by exploiting the symmetry of its limits. Both limits sit at distance $R(x)=\sqrt{0.9649^{2}-(x-0.4167)^{2}}$ from the centre $z=0.9621$, so the height of the slab is just twice that radius:

   $\int_{0.9621-R(x)}^{0.9621+R(x)}1\,dz=2R(x)=2\sqrt{0.9649^{2}-(x-0.4167)^{2}}$

   The centre $0.9621$ cancels and never appears again — only the half-width matters.

3. 3

   Do the $y$-integral, which is a plain length. The remaining integrand has no $y$ in it, so integrating over $y$ multiplies by the length of the $y$-interval:

   $L(x)=0.6773-\left(1.7476-\sqrt{1.7476^{2}-x^{2}}\right)=\sqrt{1.7476^{2}-x^{2}}-1.0703$

   Check that $L(x)>0$ across the whole $x$-range: $L(x)>0$ requires $x^{2}<1.7476^{2}-1.0703^{2}=1.90857$, i.e. $x<1.38151$, and the upper $x$-limit is $1.3815$. The region is non-degenerate right up to the edge.

4. 4

   Reduce to one integral in $x$. Combining the two results,

   $V=\int_{0.34356}^{1.3815}2\sqrt{0.9649^{2}-(x-0.4167)^{2}}\left(\sqrt{1.7476^{2}-x^{2}}-1.0703\right)dx$

   The product of two different square roots has no elementary antiderivative, so this last step has to be numerical.

5. 5

   Integrate numerically and confirm the value is stable. A midpoint rule on this integrand gives $0.73418869$ with $10^{5}$ subintervals, $0.73418869$ with $10^{6}$, and $0.73418869$ with $10^{7}$; an independent Simpson evaluation agrees to eight digits. So

   $V\approx 0.7342$

   The agreement across three refinements and two different rules is what makes this value trustworthy — a single coarse evaluation would not be.