Question:
If for some \(a \in \mathbb{R}\),
\[ \int_1^4 \int_{-x}^x \frac{1}{x^2 + y^2} dydx = \int_{-\pi/4}^{\pi/4} \int_{\sec\theta}^{a \sec\theta} \frac{1}{r} drd\theta, \]
then the value of \(a\) equals ................
If for some \(a \in \mathbb{R}\),
\[ \int_1^4 \int_{-x}^x \frac{1}{x^2 + y^2} dydx = \int_{-\pi/4}^{\pi/4} \int_{\sec\theta}^{a \sec\theta} \frac{1}{r} drd\theta, \]
then the value of \(a\) equals ................
Show Hint
When converting integrals, sketch the region of integration first. For regions bounded by lines through the origin (like \(y=mx\)) and lines perpendicular to an axis (like \(x=c\)), polar coordinates are often very effective.
Updated On: Sep 5, 2025
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Solution and Explanation
Step 1: Understanding the Concept:
The problem requires converting a double integral from Cartesian coordinates \((x,y)\) to polar coordinates \((r, \theta)\). We need to determine the limits of integration in the polar system that correspond to the given Cartesian region.
Step 2: Key Formula or Approach:
The conversion formulas from Cartesian to polar coordinates are: \[ x = r\cos\theta, \quad y = r\sin\theta, \quad x^2 + y^2 = r^2, \quad dydx = r drd\theta \] We must map the boundaries of the Cartesian region of integration to their polar equivalents.
Step 3: Detailed Explanation:
Step 3.1: Analyze the region of integration in Cartesian coordinates.
The limits are: - \(1 \le x \le 4\) - \(-x \le y \le x\) This region is bounded by the vertical lines \(x=1\) and \(x=4\), and the lines \(y=x\) and \(y=-x\). This forms a trapezoidal shape in the right half-plane.
Step 3.2: Convert the boundaries to polar coordinates.
- The line \(y=x\) corresponds to \(\tan\theta = y/x = 1\), which means \(\theta = \pi/4\).
- The line \(y=-x\) corresponds to \(\tan\theta = y/x = -1\), which means \(\theta = -\pi/4\).
So the range for \(\theta\) is \(-\pi/4 \le \theta \le \pi/4\). This matches the outer integral limits in the given polar form.
- The line \(x=1\) corresponds to \(r\cos\theta = 1\), which gives \(r = \frac{1}{\cos\theta} = \sec\theta\). This is the inner boundary for \(r\). This matches the lower limit of the inner integral.
- The line \(x=4\) corresponds to \(r\cos\theta = 4\), which gives \(r = \frac{4}{\cos\theta} = 4\sec\theta\). This is the outer boundary for \(r\).
Step 3.3: Construct the integral in polar coordinates.
The integrand \(\frac{1}{x^2+y^2}\) becomes \(\frac{1}{r^2}\). The differential area element \(dydx\) becomes \(r drd\theta\). So the integral becomes: \[ \int_{-\pi/4}^{\pi/4} \int_{\sec\theta}^{4\sec\theta} \frac{1}{r^2} . r drd\theta = \int_{-\pi/4}^{\pi/4} \int_{\sec\theta}^{4\sec\theta} \frac{1}{r} drd\theta \] This expression must be equal to the given polar integral: \[ \int_{-\pi/4}^{\pi/4} \int_{\sec\theta}^{a \sec\theta} \frac{1}{r} drd\theta \] By comparing the upper limits of the inner integral, we can conclude that: \[ a\sec\theta = 4\sec\theta \implies a = 4 \] Step 4: Final Answer:
The value of \(a\) is 4.
Step 5: Why This is Correct:
The transformation of the Cartesian boundaries into their polar counterparts was performed correctly. The resulting polar integral limits directly yield the value of \(a\) by comparison.
The problem requires converting a double integral from Cartesian coordinates \((x,y)\) to polar coordinates \((r, \theta)\). We need to determine the limits of integration in the polar system that correspond to the given Cartesian region.
Step 2: Key Formula or Approach:
The conversion formulas from Cartesian to polar coordinates are: \[ x = r\cos\theta, \quad y = r\sin\theta, \quad x^2 + y^2 = r^2, \quad dydx = r drd\theta \] We must map the boundaries of the Cartesian region of integration to their polar equivalents.
Step 3: Detailed Explanation:
Step 3.1: Analyze the region of integration in Cartesian coordinates.
The limits are: - \(1 \le x \le 4\) - \(-x \le y \le x\) This region is bounded by the vertical lines \(x=1\) and \(x=4\), and the lines \(y=x\) and \(y=-x\). This forms a trapezoidal shape in the right half-plane.
Step 3.2: Convert the boundaries to polar coordinates.
- The line \(y=x\) corresponds to \(\tan\theta = y/x = 1\), which means \(\theta = \pi/4\).
- The line \(y=-x\) corresponds to \(\tan\theta = y/x = -1\), which means \(\theta = -\pi/4\).
So the range for \(\theta\) is \(-\pi/4 \le \theta \le \pi/4\). This matches the outer integral limits in the given polar form.
- The line \(x=1\) corresponds to \(r\cos\theta = 1\), which gives \(r = \frac{1}{\cos\theta} = \sec\theta\). This is the inner boundary for \(r\). This matches the lower limit of the inner integral.
- The line \(x=4\) corresponds to \(r\cos\theta = 4\), which gives \(r = \frac{4}{\cos\theta} = 4\sec\theta\). This is the outer boundary for \(r\).
Step 3.3: Construct the integral in polar coordinates.
The integrand \(\frac{1}{x^2+y^2}\) becomes \(\frac{1}{r^2}\). The differential area element \(dydx\) becomes \(r drd\theta\). So the integral becomes: \[ \int_{-\pi/4}^{\pi/4} \int_{\sec\theta}^{4\sec\theta} \frac{1}{r^2} . r drd\theta = \int_{-\pi/4}^{\pi/4} \int_{\sec\theta}^{4\sec\theta} \frac{1}{r} drd\theta \] This expression must be equal to the given polar integral: \[ \int_{-\pi/4}^{\pi/4} \int_{\sec\theta}^{a \sec\theta} \frac{1}{r} drd\theta \] By comparing the upper limits of the inner integral, we can conclude that: \[ a\sec\theta = 4\sec\theta \implies a = 4 \] Step 4: Final Answer:
The value of \(a\) is 4.
Step 5: Why This is Correct:
The transformation of the Cartesian boundaries into their polar counterparts was performed correctly. The resulting polar integral limits directly yield the value of \(a\) by comparison.
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