Let R be the region bounded by the parabola \(y^{2} = -9x\) and the lines x = −9 and x = −4. Which of the following statements are correct?
(A) R is symmetric about the x-axis
(B) The area of R is \(\displaystyle 2\int_{-9}^{-4}3\sqrt{-x}\,dx\)
(C) The area of R is 38 sq. units
(D) The part of R above the x-axis has area 38 sq. units
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
Replacing y by −y does not change \(y^{2} = -9x\), so the parabola, and with it R, is symmetric about the x-axis. The parabola opens to the left (x ≤ 0), and its upper branch is \(y = 3\sqrt{-x}\).
The part of R above the x-axis, with the substitution u = −x (then dx = −du, and x = −9, −4 give u = 9, 4, so the limits swap):
\( \displaystyle \int_{-9}^{-4}3\sqrt{-x}\,dx = \int_{4}^{9}3\sqrt{u}\,du \)
\( \displaystyle = \left[2u^{3/2}\right]_{4}^{9} \)
\(= 2(27-8) = 38\)
The part below the x-axis is its mirror image, so the whole of R has area
\( \displaystyle 2\int_{-9}^{-4}3\sqrt{-x}\,dx = 76 \)
So R is symmetric about the x-axis, its upper part has area 38 and the whole region has area 76.
Checking the statements
(A) R is symmetric about the x-axis — Correct. \((-y)^{2} = y^{2}\).
(B) The area of R is \(\displaystyle 2\int_{-9}^{-4}3\sqrt{-x}\,dx\) — Correct. Twice the upper part, both branches counted.
(C) The area of R is 38 sq. units — Incorrect. The area of R is 76; 38 is only the upper part.
(D) The part of R above the x-axis has area 38 sq. units — Correct. \(\left[2u^{3/2}\right]_{4}^{9} = 54-16 = 38\).