Find the area bounded by the following curves: y=frac3x
Find the area bounded by the following curves: $$ y=\frac3x, \qquad x^2+y^2=10. $$
Explanation of Method
Solving the hyperbola and the circle simultaneously reduces to a biquadratic equation, giving four intersection points — two in the first quadrant and, by the odd symmetry of $y=3/x$ and the full rotational symmetry of the circle, two symmetric points in the third quadrant. The bounded region between the branches thus consists of two congruent pieces (first and third quadrant), and the first-quadrant piece is computed via a circular-segment antiderivative combined with an elementary logarithm.
- Find the intersection points by substitution, reducing to a quadratic in $x^2$.
- Set up the first-quadrant area integral and double it by symmetry.
- Evaluate using the circular-segment antiderivative and a logarithm.
Step 1 · Find the intersection points
Substituting $y=\dfrac3x$ into $x^2+y^2=10$:
$$ x^2+\frac9{x^2}=10 \quad\Longrightarrow\quad x^4-10x^2+9=0. $$
With $u=x^2$: $u^2-10u+9=0\Rightarrow(u-1)(u-9)=0\Rightarrow u=1$ or $u=9$, so $x=\pm1$ or $x=\pm3$. In the first quadrant: $(1,3)$ and $(3,1)$; by the symmetry $(x,y)\to(-x,-y)$ of both curves, the third quadrant contributes the mirror points $(-1,-3)$ and $(-3,-1)$.
Step 2 · Set up the first-quadrant area, doubled by symmetry
For $1\leqslant x\leqslant3$, the circle $y=\sqrt{10-x^2}$ lies above the hyperbola $y=3/x$ (checked, e.g., at $x=\sqrt3$: circle gives $\sqrt7\approx2.65$, hyperbola gives $\sqrt3\approx1.73$). The total area is twice the first-quadrant contribution:
$$ A = 2\int_1^3\left[\sqrt{10-x^2}-\frac3x\right]dx. $$
Step 3 · Evaluate each piece
Using $\displaystyle\int\sqrt{a^2-x^2}\,dx=\frac x2\sqrt{a^2-x^2}+\frac{a^2}2\arcsin\frac xa+C$ with $a=\sqrt{10}$:
$$ \int_1^3\sqrt{10-x^2}\,dx = \left[\frac x2\sqrt{10-x^2}+5\arcsin\frac x{\sqrt{10}}\right]_1^3. $$
Since $\frac x2\sqrt{10-x^2}$ equals $\frac32$ at both $x=1$ (giving $\frac12\sqrt9=\frac32$) and $x=3$ (giving $\frac32\sqrt1=\frac32$), that term cancels in the difference, leaving:
$$ \int_1^3\sqrt{10-x^2}\,dx = 5\left[\arcsin\frac3{\sqrt{10}}-\arcsin\frac1{\sqrt{10}}\right]. $$
For the hyperbola term:
$$ \int_1^3\frac3x\,dx = 3\big[\ln x\big]_1^3 = 3\ln3. $$
Step 4 · Combine
$$ A = 2\left(5\left[\arcsin\frac3{\sqrt{10}}-\arcsin\frac1{\sqrt{10}}\right]-3\ln3\right) = 10\left[\arcsin\frac3{\sqrt{10}}-\arcsin\frac1{\sqrt{10}}\right]-6\ln3. $$
Additional Notes
- Verified numerically: $10\left[\arcsin\frac3{\sqrt{10}}-\arcsin\frac1{\sqrt{10}}\right]-6\ln3\approx2.68128$, matching direct numerical integration of the first-quadrant piece (doubled) to $6$ significant figures.
- Since $\arcsin\frac3{\sqrt{10}}$ and $\arcsin\frac1{\sqrt{10}}$ are complementary in the sense that $\frac3{\sqrt{10}}$ and $\frac1{\sqrt{10}}$ are the sine and cosine of the same acute angle (as $3^2+1^2=10$), the bracket can also be written as $\arcsin\frac3{\sqrt{10}}-\arccos\frac3{\sqrt{10}}$, though the form above is simplest to evaluate directly.
Result
The area is: