Question
The area bounded by the curves $y=-\sqrt{-x}$ and $x=-\sqrt{-y}$ where $x, y \leq 0$(a) cannot be determined(b) is $1 / 3$(c) is $2 / 3$(d) is same as that of the figure bounded by the curves $y=\sqrt{-x} ; x \leq 0$ and $x=\sqrt{-y} ; y \leq 0$
Step 1
Squaring both sides of the equations, we get $y^2=-x$ and $x^2=-y$ respectively. Show more…
Show all steps
Your feedback will help us improve your experience
Vikash Ranjan and 86 other Calculus 2 / BC educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The area bounded by the curves $y=\sqrt{x}, x=2 y+3$ in the first quadrant and $x$ -axis is (a) $2 \sqrt{3}$ (b) 18 (c) 9. (d) $18 \sqrt{3}$
The area bounded by the lines $y=2, x=1, x=a$ and the curve $y=f(x)$, which cuts the last two lines above the first line for all $a \geq 1$, is equal to $\cdot \frac{2}{3}\left[(2 a)^{3 / 2}-3 a+3-2 \sqrt{2}\right]$ Then $f(x)=$ (A) $2 \sqrt{2 x} x \geq 1$ (B) $\sqrt{2 x}, x \geq 1$ (C) $2 \sqrt{x}, x \geq 1$ (D) None of these
The area bounded by the lines $y=2, x=1, x=a$ and the curve $y=f(x)$, which cuts the last two lines above the first line for all $a \geq 1$, is equal to $\frac{2}{3}\left[(2 a)^{3 / 2}-3 a+3-2 \sqrt{2}\right]$. Then, $f(x)=$ (A) $2 \sqrt{2 x} x \geq 1$ (B) $\sqrt{2 x}, x \geq 1$ (C) $2 \sqrt{x}, x \geq 1$ (D) None of these
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD