What is the area of the smaller part of the…

What is the area of the smaller part of the circle, to the right of x = 3?

Question 51

HardOriginal question

The circle $x^{2} + y^{2} = 25$ is cut by the line $x = 3$ into a smaller and a larger part.

What is the area of the smaller part of the circle, to the right of $x = 3$ ?

- \frac{25 \sin^{-1}{\left(\frac{3}{5} \right)}}{2} - 6 + \frac{25 \pi}{4}

square unitsCorrect answerYour answer8 square unitsCorrect answerYour answer

- 25 \sin^{-1}{\left(\frac{3}{5} \right)} - 12 + \frac{25 \pi}{2}

square unitsCorrect answerYour answer

12 + 25 \sin^{-1}{\left(\frac{3}{5} \right)} + \frac{25 \pi}{2}

square unitsCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$2\int_3^{5} \sqrt{25 - x^{2}}\,dx$ , using $\int \sqrt{25 - x^{2}}\,dx = \frac{x}{2}\sqrt{25 - x^{2}} + \frac{25}{2}\sin^{-1}\frac{x}{5}$ .
Between 3 and 5: $\frac{25\pi}{4} - 6 - \frac{25}{2}\sin^{-1}\frac{3}{5}$ ; doubled, $- 25 \sin^{-1}{\left(\frac{3}{5} \right)} - 12 + \frac{25 \pi}{2}$ .

Worth noting

Half of this is the part above the $x$ -axis only.

Ten questions from Area of Bounded Regions, at your own pace.

Ten questions from Area of Bounded Regions, at your own pace.