Q17Mathematics I

The mathematical problem requires the exact step-by-step evaluation of the 3D volumetric integral with dynamic variable bounds:

Step 1: Inner Integration with respect to $x$

We begin strictly with the innermost integral, integrating exclusively with respect to $x$. Both $y$ and $z$ are treated rigidly as constants. We isolate the $x$ component:

We carefully multiply this result back into the remaining integrand constants ($yz$):

Step 2: Middle Integration with respect to $y$

Next, we focus solely on the middle integral, integrating strictly with respect to $y$, while treating $z$ as a constant. We can factor out the $\frac{z^3}{2}$ term:

Applying the exact upper and lower bounds ($z$ and $1$):

The overall integral formula is now reduced to a single-variable calculation:

Step 3: Outer Integration with respect to $z$

Finally, we evaluate this standard definite integral along the fixed numerical bounds from $1$ to $2$. The constant factor $1/8$ is pulled entirely out of the integral:

We carefully compute the numerical value at the upper limit ($z=2$):

And at the lower limit ($z=1$):

Applying the Fundamental Theorem of Calculus (Upper - Lower):

The fraction $\frac{225}{64}$ is exactly $3.515625$. Let me re-read the target solution. The provided solution summary stated the answer is $2.25$ (which is $9/4$). My calculation yields $225/64$. Why the discrepancy? Let's re-read the provided solution summary equation exactly: ${\int }_1^2{\int }_1^{z}yz{\left[\frac{x^2}{2}\right]}_0^{yz}dydz={\int }_1^2{\int }_1^z\frac{y^3{z}^3}{2}dydz=2.25$. The integral of $y^3{z}^3/2$ dy is $\frac{z^3}{2}[y^4/4{]}_1^z=\frac{z^3}{8}(z^4-1)=\frac{z^7-z^3}{8}$. The integral of this dz from 1 to 2 is $\frac{1}{8}[z^8/8-z^4/4{]}_1^2=\frac{1}{8}[(256/8-16/4)-(1/8-1/4)]=\frac{1}{8}[28-(-1/8)]=225/64=3.515625$. If the answer was $2.25$, then $2.25=9/4=144/64$. Ah! Let me check the bounds again. If the inner integral limit was $x$ from $0$ to $1$ or something? No, the summary says $[x^2/2{]}_0^{yz}$. Wait! What if the integrand was just $1$ instead of $xyz$? ${\int }_0^{yz}1dx=yz$. Then $\int yzdy=y^{2}z/2$, etc. What if the integrand was $x$? ${\int }_0^{yz}xdx=y^2{z}^2/2$. But the summary says yz [x^2/2]_0^{yz} which means the integrand was $xyz$. Let me just explicitly calculate it again. $225/64$ is correct math for the stated integral in the summary. I will write the explanation to arrive at $225/64$ and state it correctly, since $2.25$ in the summary must be an arithmetic error in the source material's summary.

The exact mathematical result is $225/64$ (or $3.515625$).