Find the volume of the solid bounded by the paraboloid
Find the volume of the solid bounded by the paraboloid of revolution: $$ z=x^2+y^2 $$ and the planes: $$ z=0,\quad y=1,\quad y=2x,\quad y=6-x. $$
Explanation of Method
The base region is the triangle cut out by $y=1$, $y=2x$, and $y=6-x$; it is more convenient to slice horizontally (integrating in $x$ first, for each fixed $y$), since the left and right boundaries $x=y/2$ and $x=6-y$ are each single-valued functions of $y$ over the triangle's full height.
- Find the triangle's vertices and the range of $y$.
- Set up the volume integral, slicing horizontally.
- Evaluate the resulting integral in $y$.
Step 1 · Find the triangle's vertices
Intersecting $y=1$ with $y=2x$ gives $\left(\frac12,1\right)$; intersecting $y=1$ with $y=6-x$ gives $(5,1)$; intersecting $y=2x$ with $y=6-x$ gives $x=2,\,y=4$, i.e. $(2,4)$. So $y$ ranges from $1$ to $4$, with $x$ running from $\frac y2$ (on $y=2x$) to $6-y$ (on $y=6-x$).
Step 2 · Set up the volume integral
$$ V = \int_1^4\!\int_{y/2}^{6-y}\big(x^2+y^2\big)\,dx\,dy. $$
Step 3 · Evaluate the inner integral
$$ \int_{y/2}^{6-y}(x^2+y^2)\,dx = \left[\frac{x^3}3+xy^2\right]_{y/2}^{6-y} = \left[\frac{(6-y)^3}3+(6-y)y^2\right]-\left[\frac{y^3}{24}+\frac{y^3}2\right]. $$
Expanding $(6-y)^3=216-108y+18y^2-y^3$ and combining:
$$ \frac{(6-y)^3}3+(6-y)y^2 = 72-36y+12y^2-\frac{4y^3}3, $$
and subtracting $\dfrac{y^3}{24}+\dfrac{y^3}2=\dfrac{13y^3}{24}$:
$$ \int_{y/2}^{6-y}(x^2+y^2)\,dx = 72-36y+12y^2-\frac{4y^3}3-\frac{13y^3}{24} = 72-36y+12y^2-\frac{15y^3}8. $$
Step 4 · Evaluate the outer integral
$$ V = \int_1^4\left(72-36y+12y^2-\frac{15y^3}8\right)dy = \left[72y-18y^2+4y^3-\frac{15y^4}{32}\right]_1^4. $$
At $y=4$: $288-288+256-\dfrac{3840}{32}=256-120=136$. At $y=1$: $72-18+4-\dfrac{15}{32}=58-\dfrac{15}{32}=\dfrac{1841}{32}$. So:
$$ V = 136-\frac{1841}{32} = \frac{4352-1841}{32} = \frac{2511}{32}. $$
Additional Notes
- This derivation matches the source exactly.
- Verified symbolically: direct evaluation of the nested integral gives exactly $\dfrac{2511}{32}\approx78.469$.
Result
The volume is: