Find the volume of the solid bounded by the surface: z=x
Find the volume of the solid bounded by the surface: $$ z=x\sqrt x+y\sqrt y, $$ the plane $x+y=1$, and $x>0,\,y>0,\,z>0$.
Explanation of Method
A note on the source's interpretation: the literal boundary equation was transcribed in the source as "$x+y=0$," which together with $x,y>0$ describes only the single point (origin) — giving no volume at all. The source itself recognized this and adopted $x+y=1$ instead, which gives the natural triangular base region; that reading is adopted here.
Over the triangle $x,y\geqslant0,\,x+y\leqslant1$, the height $z=x^{3/2}+y^{3/2}$ is a direct sum, so the double integral separates into two identical-structure single integrals via Fubini's theorem.
- Set up the volume as $\displaystyle\iint_D(x^{3/2}+y^{3/2})\,dA$ over the triangle $D$.
- Integrate in $y$ first, then in $x$.
- Evaluate each resulting integral, optionally via the Beta function.
Step 1 · Set up the volume integral
$$ V = \int_0^1\int_0^{1-x}\left(x^{3/2}+y^{3/2}\right)dy\,dx. $$
Step 2 · Evaluate the inner integral
$$ \int_0^{1-x}\left(x^{3/2}+y^{3/2}\right)dy = x^{3/2}(1-x)+\frac25(1-x)^{5/2}. $$
Step 3 · Evaluate the outer integral
$$ I_1=\int_0^1x^{3/2}(1-x)\,dx = \int_0^1\left(x^{3/2}-x^{5/2}\right)dx = \frac25-\frac27 = \frac4{35}. $$
$$ I_2=\int_0^1(1-x)^{5/2}\,dx = \int_0^1u^{5/2}\,du = \frac27\qquad(u=1-x). $$
$$ V = I_1+\frac25I_2 = \frac4{35}+\frac25\cdot\frac27 = \frac4{35}+\frac4{35} = \frac8{35}. $$
Additional Notes
- This derivation matches the source's own corrected reading and final answer.
- Verified symbolically (including cross-checking via the Beta function, $\int_0^1x^{3/2}(1-x)\,dx=B\left(\frac52,2\right)=\frac4{35}$ and $\int_0^1(1-x)^{5/2}\,dx=B\left(1,\frac72\right)=\frac27$) and by direct numerical double integration, both giving exactly $\frac8{35}$.
Result
The volume is: