Question

Which integral represents the enclosed area of the ellipse after using the positive first-quadrant value of \[y\]?

पर्याय

• \[\int_0^a \frac{b}{a}\sqrt{x^2-a^2}\,dx\]

• \[\left|\int_0^a \frac{b}{a}\sqrt{a^2-x^2}\,dx\right|\]

• \[4\int_0^a \frac{b}{a}\sqrt{a^2-x^2}\,dx\]

• \[ 4\int_0^b \frac{b}{a}\sqrt{b^2 - y^2}\, dy\]

Answer 1

In the first quadrant, \[y=\frac{b}{a}\sqrt{a^2-x^2}\]. The area in that quadrant is multiplied by \[4\] because the ellipse is symmetrical about both axes.