1. **State the problem:** We need to evaluate the double integral $$\iint_R x^2 \, dA$$ where the region $R$ is in the first quadrant enclosed by the curves $$xy=1,$$ $$y=x,$$ and $$y=2x.$$ 2. **Understand the region:** The region $R$ is bounded by three curves: - Hyperbola: $$xy=1 \implies y=\frac{1}{x}$$ - Line 1: $$y=x$$ - Line 2: $$y=2x$$ Since $R$ is in the first quadrant, $x>0$ and $y>0$. 3. **Find the intersection points of the curves to determine limits:** - Intersection of $$y=x$$ and $$xy=1$$: Substitute $y=x$ into $xy=1$ gives $$x \cdot x = 1 \implies x^2=1 \implies x=1$$ (only positive root since first quadrant). So intersection point is $(1,1)$. - Intersection of $$y=2x$$ and $$xy=1$$: Substitute $y=2x$ into $xy=1$ gives $$x \cdot 2x = 1 \implies 2x^2=1 \implies x^2=\frac{1}{2} \implies x=\frac{1}{\sqrt{2}}$$. So intersection point is $$\left(\frac{1}{\sqrt{2}}, \frac{2}{\sqrt{2}}\right) = \left(\frac{1}{\sqrt{2}}, \sqrt{2}\right)$$. - Intersection of lines $$y=x$$ and $$y=2x$$ is at $x=0$, but since $x>0$, they meet only at the origin which is outside the region bounded by $xy=1$. 4. **Set up the integral:** The region $R$ is bounded between the two lines $y=x$ and $y=2x$, and above the hyperbola $xy=1$. For a fixed $x$, $y$ varies between $y=x$ and $y=2x$. But the hyperbola $xy=1$ can be rewritten as $y=\frac{1}{x}$. Since the region is enclosed by these curves, the $x$-values range from $$x=\frac{1}{\sqrt{2}}$$ to $$x=1$$ (from the intersection points found). For each $x$ in this interval, $y$ varies from the hyperbola $y=\frac{1}{x}$ up to the line $y=2x$ or $y=x$ depending on which is lower or higher. Check which line is above the hyperbola in this interval: - At $x=\frac{1}{\sqrt{2}}$, $y=\frac{1}{x} = \sqrt{2} \approx 1.414$. - $y=x = \frac{1}{\sqrt{2}} \approx 0.707$. - $y=2x = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.414$. At $x=\frac{1}{\sqrt{2}}$, $y=\frac{1}{x} = y=2x$. At $x=1$, $y=\frac{1}{1} =1$, $y=x=1$, $y=2x=2$. So between $x=\frac{1}{\sqrt{2}}$ and $x=1$, the region is bounded below by $y=\frac{1}{x}$ and above by $y=x$ for the left part, and by $y=2x$ for the right part. But the problem states the region is enclosed by $xy=1$, $y=x$, and $y=2x$. Hence, the region is between $y=x$ and $y=2x$, and above $xy=1$. Therefore, for $x$ in $\left[\frac{1}{\sqrt{2}}, 1\right]$, $y$ varies from $y=\frac{1}{x}$ to $y=2x$. For $x$ in $\left[1, 2\right]$, $y$ varies from $y=\frac{1}{x}$ to $y=x$. But since the intersection of $y=x$ and $xy=1$ is at $x=1$, and $y=2x$ and $xy=1$ is at $x=\frac{1}{\sqrt{2}}$, the region is between $x=\frac{1}{\sqrt{2}}$ and $x=1$ bounded by $y=\frac{1}{x}$ and $y=2x$, and between $x=1$ and $x=2$ bounded by $y=\frac{1}{x}$ and $y=x$. 5. **Split the integral into two parts:** $$\iint_R x^2 \, dA = \int_{x=\frac{1}{\sqrt{2}}}^1 \int_{y=\frac{1}{x}}^{2x} x^2 \, dy \, dx + \int_1^2 \int_{y=\frac{1}{x}}^{x} x^2 \, dy \, dx$$ 6. **Evaluate the inner integrals:** Since $x^2$ is constant with respect to $y$, $$\int_{y=a}^{b} x^2 \, dy = x^2 (b - a)$$ So, First integral: $$\int_{x=\frac{1}{\sqrt{2}}}^1 x^2 \left(2x - \frac{1}{x}\right) dx = \int_{\frac{1}{\sqrt{2}}}^1 x^2 \left(2x - \frac{1}{x}\right) dx = \int_{\frac{1}{\sqrt{2}}}^1 (2x^3 - x) dx$$ Second integral: $$\int_1^2 x^2 \left(x - \frac{1}{x}\right) dx = \int_1^2 x^2 \left(x - \frac{1}{x}\right) dx = \int_1^2 (x^3 - x) dx$$ 7. **Calculate the first integral:** $$\int_{\frac{1}{\sqrt{2}}}^1 (2x^3 - x) dx = \left[ \frac{2x^4}{4} - \frac{x^2}{2} \right]_{\frac{1}{\sqrt{2}}}^1 = \left[ \frac{x^4}{2} - \frac{x^2}{2} \right]_{\frac{1}{\sqrt{2}}}^1$$ Evaluate at $x=1$: $$\frac{1^4}{2} - \frac{1^2}{2} = \frac{1}{2} - \frac{1}{2} = 0$$ Evaluate at $x=\frac{1}{\sqrt{2}}$: $$\frac{\left(\frac{1}{\sqrt{2}}\right)^4}{2} - \frac{\left(\frac{1}{\sqrt{2}}\right)^2}{2} = \frac{\frac{1}{4}}{2} - \frac{\frac{1}{2}}{2} = \frac{1}{8} - \frac{1}{4} = -\frac{1}{8}$$ So the first integral is: $$0 - \left(-\frac{1}{8}\right) = \frac{1}{8}$$ 8. **Calculate the second integral:** $$\int_1^2 (x^3 - x) dx = \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_1^2$$ Evaluate at $x=2$: $$\frac{2^4}{4} - \frac{2^2}{2} = \frac{16}{4} - \frac{4}{2} = 4 - 2 = 2$$ Evaluate at $x=1$: $$\frac{1^4}{4} - \frac{1^2}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$$ So the second integral is: $$2 - \left(-\frac{1}{4}\right) = 2 + \frac{1}{4} = \frac{9}{4}$$ 9. **Add both results:** $$\frac{1}{8} + \frac{9}{4} = \frac{1}{8} + \frac{18}{8} = \frac{19}{8}$$ **Final answer:** $$\boxed{\frac{19}{8}}$$ --- "slug": "double integral", "subject": "calculus", "desmos": {"latex": "y=x, y=2x, xy=1", "features": {"intercepts": true, "extrema": true}}, "q_count": 1