More Volume Problems — Question 7

Question 7

The base of a solid is the region bounded by $$y=\ln(x+1) \quad\text{and}\quad y=\frac{x}{2},$$ between their points of intersection. Cross-sections perpendicular to the $x$-axis are right triangles whose height is equal to the base. Find the volume of the solid.

Question 7 – Solution

1. Define the intersection bounds.

Let $b(x)=\ln(1+x)-x/2$. Since $b'(x)=(1-x)/(2(1+x))$, $b$ increases to $x=1$ and then decreases. Its zeros are $0$ and a unique $a>1$ with$$\ln(1+a)=a/2,\qquad a\approx2.512862417.$$This equation defines $a$ exactly; $b(x)>0$ on $(0,a)$.

2. Find the triangular area and set up the volume.

Both perpendicular legs have length $b(x)$, so$$V=\frac12\int_0^a\left(\ln(1+x)-\frac x2\right)^2dx.$$

3. Expand and find the needed antiderivatives.

Write $L=\ln(1+x)$ and $u=1+x$. The square is $L^2-xL+x^2/4$. Integration by parts gives$$\int L^2dx=u(L^2-2L+2),\qquad \int xL\,dx=\frac{x^2}{2}L-\frac{x^2}{4}+\frac x2-\frac L2.$$For the first identity, use $\int\ln^2u\,du=u\ln^2u-2\int\ln u\,du$.

4. Combine the terms and apply the endpoints.

An antiderivative of the squared expression is$$H(x)=uL^2-\left(u+\frac{u^2}{2}\right)L+u+\frac{u^2}{4}+\frac{x^3}{12}.$$At $x=0$, $H(0)=5/4$. Substituting $u=1+a$ and $L=a/2$ gives$$H(a)-H(0)=\frac{a^3}{12}-\frac{a^2}{2}+\frac{3a}{4}=\frac{a(a-3)^2}{12}.$$

5. Include the triangular-area factor.

$$V=\frac12[H(a)-H(0)]=\boxed{\frac{a(a-3)^2}{24}}\approx0.024846244.$$

All volumes are in cubic units.