0

enter image description here

I am supposed to find the area of the blue shaded region. The line $\mathrm {OA}$ is $y=x$. The circle is $x^2+y^2=16$.

The best method is to find slope of the line and use the formula $\frac12r^2\theta$ for the area of sector. But I'm supposed to use application of definite integrals.

In the figure, $A\equiv(2\sqrt2,2\sqrt2)$, $B\equiv(2\sqrt2,0)$, $C\equiv(4,0)$.

The area of the triangle can be found out by $\frac12(AB)(OB)$. To find area of blue part of $ABC$, I applied the definite integral $$\int _{2\sqrt2} ^4 \sqrt{16-x^2}dx$$

Now, won't this integral include the red shaded region too? Am I supposed to divide the final answer by two, so as to get only the blue shaded region? Because if I do divide, the answer doesn't match with what I get from $\frac12r^2\theta$.

Tejas
  • 2,082

1 Answers1

1

No, the integral of the positive square root function over that interval only measures the remaining blue area. (Not including the triangle)

$$\begin{align} \int_{2\sqrt 2}^4\; {+}\sqrt{16-x^2}\operatorname d x & = \int_{\pi/4}^0 4 \sin \theta \times (- 4 \sin (\theta) )\operatorname d \theta \quad \text{by substituting }x=4\cos \theta \\ & = 8 \int_0^{\pi/4} (1-\cos 2 \theta) \operatorname d \theta \\ & = \left[8\theta - 4\sin 2\theta \right]_{\theta = 0}^{\theta = \pi/4} \\ & = 2\pi - 4 \end{align} $$

Then add the area of the triangle to get the final answer: $2\pi\color{gray}{= \tfrac 1 2 (16)^2 \tfrac {\pi} 4}$

Graham Kemp
  • 129,094
  • Shouldn't the upper limit of the integral be $4$? And the final answer should be $2\pi-4$. The area of the triangle is $4$. So, we get overall area $2\pi$. Thanks a lot! :-) – Tejas Jan 29 '15 at 04:21