Question:
What is the value of the integral \(\iint_D dA\), where D is the region bounded by the curve \(y=x^2\) and the line \(y=4\)?
What is the value of the integral \(\iint_D dA\), where D is the region bounded by the curve \(y=x^2\) and the line \(y=4\)?
Show Hint
Find where the parabola meets the line, then integrate the vertical strip width over that range.
Updated On: Jul 3, 2026
- 8/3
- 11/3
- 16/3
- 32/3
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The Correct Option is D
Solution and Explanation
Step 1: Find where the curve and the line meet. Set \(x^2=4\), which gives \(x=-2\) and \(x=2\). So the region D is bounded above by \(y=4\) and below by \(y=x^2\), for \(-2\le x\le 2\).
Step 2: Set up the double integral as a single integral over x. \[\iint_D dA=\int_{-2}^{2}\int_{x^2}^{4} dy\, dx=\int_{-2}^{2}(4-x^2)\,dx.\]
Step 3: Evaluate. \[\int_{-2}^{2}(4-x^2)\,dx=\left[4x-\frac{x^3}{3}\right]_{-2}^{2}=\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)=16-\frac{16}{3}=\frac{48-16}{3}=\frac{32}{3}.\] \[\boxed{\iint_D dA=\frac{32}{3}}\]
Step 2: Set up the double integral as a single integral over x. \[\iint_D dA=\int_{-2}^{2}\int_{x^2}^{4} dy\, dx=\int_{-2}^{2}(4-x^2)\,dx.\]
Step 3: Evaluate. \[\int_{-2}^{2}(4-x^2)\,dx=\left[4x-\frac{x^3}{3}\right]_{-2}^{2}=\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)=16-\frac{16}{3}=\frac{48-16}{3}=\frac{32}{3}.\] \[\boxed{\iint_D dA=\frac{32}{3}}\]
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