How do you solve the curve equation $y^2 = 2x$ for y to use in the integral setup?

Conditions

• The boundary curve is given as $y^2 = 2x$.
• The region is in the first quadrant ($x \ge 0, y \ge 0$).
• Vertical strips are being used, requiring height as a function of $x$.

Reasoning, step by step

1. Start with the equation $y^2 = 2x$.
2. Take the square root of both sides: $y = \pm\sqrt{2x}$.
3. Simplify the radical: $y = \pm\sqrt{2} \sqrt{x} = \pm\sqrt{2} x^{1/2}$.
4. Determine the correct sign based on the location of the region. For the first quadrant, $y$ is positive.
5. Select the positive root: $y = \sqrt{2} x^{1/2}$.
6. Use this expression for $y$ in the height calculation of the differential area element $dA$.

Example

The instructor says, "if that equation was just solved for y instead of in y squared... y is equal to square root of 2 times x to the one half." This solved equation is then substituted into the integrand.

Common misconceptions

• Trying to use $y^2 = 2x$ directly without isolating $y$, making it difficult to express the strip height purely in terms of $x$.
• Forgetting to consider the domain and selecting the negative root when the region is in the first quadrant.
• Incorrectly simplifying $\sqrt{2x}$ as $2\sqrt{x}$ instead of $\sqrt{2}x^{1/2}$.

Watch the explanation

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