Find the area bounded by the following curves (a>0)
Find the area bounded by the following curves ($a>0$): $$ xy=a^2, \qquad x+y=\frac52a. $$
Explanation of Method
Solving the hyperbola and the line simultaneously (via a quadratic in $x$) locates the two intersection points, between which the straight line lies above the hyperbola. The area is then a straightforward integral of the vertical gap between the two curves, involving an elementary logarithmic antiderivative from the $a^2/x$ term.
- Find the intersection points of the line and the hyperbola.
- Determine which curve lies above the other between the intersections.
- Integrate the vertical gap between the curves.
Step 1 · Find the intersection points
Substituting $y=\dfrac{a^2}x$ into $x+y=\dfrac52a$:
$$ x+\frac{a^2}x=\frac52a \quad\Longrightarrow\quad 2x^2-5ax+2a^2=0. $$
By the quadratic formula:
$$ x = \frac{5a\pm\sqrt{25a^2-16a^2}}4 = \frac{5a\pm3a}4 \quad\Longrightarrow\quad x=2a \ \text{ or }\ x=\frac a2. $$
Correspondingly $y=\dfrac a2$ (when $x=2a$) and $y=2a$ (when $x=\dfrac a2$): intersection points $\left(\dfrac a2,2a\right)$ and $(2a,\dfrac a2)$.
Step 2 · Determine which curve is above
For $\dfrac a2\leqslant x\leqslant2a$, check at $x=a$: the line gives $y=\dfrac52a-a=\dfrac32a$, while the hyperbola gives $y=\dfrac{a^2}a=a$. Since $\dfrac32a>a$, the line lies above the hyperbola throughout this interval.
Step 3 · Set up the area integral
$$ A = \int_{a/2}^{2a}\left[\left(\frac52a-x\right)-\frac{a^2}x\right]dx. $$
Step 4 · Evaluate the integral
$$ A = \left[\frac52ax-\frac{x^2}2-a^2\ln x\right]_{a/2}^{2a}. $$
At $x=2a$:
$$ \frac52a(2a)-\frac{(2a)^2}2-a^2\ln(2a) = 5a^2-2a^2-a^2\ln(2a) = a^2\big(3-\ln(2a)\big). $$
At $x=\dfrac a2$:
$$ \frac52a\cdot\frac a2-\frac{(a/2)^2}2-a^2\ln\frac a2 = \frac{5a^2}4-\frac{a^2}8-a^2\ln\frac a2 = a^2\left(\frac98-\ln\frac a2\right). $$
Subtracting:
$$ A = a^2\left[\left(3-\ln(2a)\right)-\left(\frac98-\ln\frac a2\right)\right] = a^2\left[\frac{15}8+\ln\frac{a/2}{2a}\right] = a^2\left[\frac{15}8+\ln\frac14\right] = a^2\left(\frac{15}8-2\ln2\right). $$
Additional Notes
- Verified numerically (at $a=1$): direct quadrature of $\displaystyle\int_{1/2}^{2}\left[\left(\frac52-x\right)-\frac1x\right]dx$ gives $\approx0.488706$, matching $\frac{15}8-2\ln2\approx0.488706$.
- Both $\ln$ terms in Step 4 combine into a single $\ln\frac14=-2\ln2$, since $\dfrac{a/2}{2a}=\dfrac14$ regardless of the value of $a$ — the answer's dependence on $a$ is therefore purely the overall scale factor $a^2$, as expected by dimensional analysis (both curves scale linearly with $a$, so enclosed areas scale as $a^2$).
Result
The area is: