CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Using integration find the area of the following region $$(x, y):|x+2|\leq y\leq \sqrt{20-x^2}$$.


Solution

$$y \ge |x + 2|$$
$$P, Q, R$$ are points of intersection $$y = \sqrt{20 - x^2}$$ with $$y = x + 2$$ & $$x = 0$$
from graph
$$P = (2, 4)$$
$$Q = (0, 2)$$
$$R = (0, 2 \sqrt{5})$$
area of shaded region $$= \displaystyle \int_0^2 (\sqrt{20 - x^2} - (x + 2)0 dx$$
$$= \displaystyle \int_0^2 \sqrt{20 - x^2} dx - \int^2_0 (x + 2) dx$$
$$= \left[\dfrac{x}{2} \sqrt{20 - x^2} + 10 \sin^{-1} \dfrac{x}{2 \sqrt{5}} \right]^2_0 - \left[\dfrac{x^2}{2} + 2x \right]^2_0$$
$$= 4 + 10 \sin^{-1} \left(\dfrac{1}{\sqrt{5}} \right) - 6$$
Area $$= 10 \sin^{-1} \left(\dfrac{1}{\sqrt{5}} \right) - 2$$

1340153_1083071_ans_700b400738694ce2b43e3ca0231fa52a.png

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image