Find the volume of the solid bounded by the planes: x=0
Find the volume of the solid bounded by the planes: $$ x=0,\quad y=0,\quad z=0,\quad y=4 $$ and the paraboloid: $$ z=x^2+y^2+1. $$
Explanation of Method
A note on the problem statement: as given, the only bound restricting $x$ from above is missing — $x=0$ provides a lower bound, but no plane like "$x=p$" is listed to close off the region in the $x$-direction, so the literal solid (with $x$ ranging over all of $[0,\infty)$) has infinite volume. The natural, most likely intended reading — matching the symmetric bound $y=4$ already given, and the fact that a $4$th vertical plane parallel to $y=4$ is exactly the kind of bound one would expect to complete this problem — is $x=4$ as well; the solution below proceeds with the general bound $x=p$ made explicit, so the formula can be evaluated for whatever value was intended, with $p=4$ given as the most likely completion.
- Set up the volume integral over the rectangle $0\leqslant x\leqslant p,\,0\leqslant y\leqslant4$, for a general upper bound $p$.
- Evaluate the double integral.
- Specialize to $p=4$ for the most likely intended answer.
Step 1 · Set up the volume integral
With $z$ running from $0$ (the plane $z=0$) up to the paraboloid $z=x^2+y^2+1$, over the rectangle $0\leqslant x\leqslant p,\,0\leqslant y\leqslant4$:
$$ V(p) = \int_0^p\!\int_0^4\big(x^2+y^2+1\big)\,dy\,dx. $$
Step 2 · Evaluate the inner integral
$$ \int_0^4(x^2+y^2+1)\,dy = \left[x^2y+\frac{y^3}3+y\right]_0^4 = 4x^2+\frac{64}3+4 = 4x^2+\frac{76}3. $$
Step 3 · Evaluate the outer integral
$$ V(p) = \int_0^p\left(4x^2+\frac{76}3\right)dx = \frac{4p^3}3+\frac{76p}3 = \frac{4p\big(p^2+19\big)}3. $$
Step 4 · Specialize to $p=4$
$$ V(4) = \frac{4\cdot4\,(16+19)}3 = \frac{16\cdot35}3 = \frac{560}3. $$
Additional Notes
- As noted above, the problem statement omits the bounding plane needed to close off the $x$-direction. The source reached the same conclusion, presenting both $p=4$ (giving $\frac{560}3$) and $p=2$ (giving $\frac{184}3$) as possible completions, and preferring $p=4$ "matching the $y$ range" as the most likely intended reading; the general formula $V(p)=\frac{4p(p^2+19)}3$ above lets the volume be recovered for whichever bound was actually intended.
- Verified symbolically: $V(p)=\frac{4p^3}3+\frac{76p}3$, and $V(4)=\frac{256}3+\frac{304}3=\frac{560}3$, matching the source's preferred value exactly.
Result
For a general upper bound $x=p$, the volume is: