Find the area of the region{(x,y): y^2 is less than equal to 6 ax and x^2 + y^2 is lessor - Maths - Application of Integrals

Question

Find the area of the region{(x,y): y^2 is less than equal to 6
ax and x^2 + y^2 is lessor equal to 16 a^2 using method of
integration

Priyanka Kedia · Accepted Answer

The given inequation be as,x2+y2≤16a2 and y2≤6axConsider the equation x2+y2=16a2 and y2=6ax corrsponding to given inequation
Now, graph the inequation, and then we have to find the shaded area
<img src=
"https://s3mn%20mnimgs%20com/img/shared/content_ck_images/ana_qa_image_57b2eb2c77f8b%20png">

To find the intersection point, eliminate y from both the equation, we get,x2+6ax=16a2 ⇒x2+6ax-16a2=0⇒x+8ax-2a=0⇒x=2a and x=-8aThus,Area=2∫02a6axdx+∫2a4a16a2-x2dx=226ax32302a+x216a2-x2+16a22sin-1x4a2a4a=226a2a323+4a216a2-16a2+8a2sin-14a4a-2a216a2-4a2-8a2sin-12a4a=24a12a23+8a2sin-11-2a212a2-8a2sin-112=28a233+8a2×π2-a12a2-8a2×π6=28a233-2a23+8a2×π2-8a2×π6=22a233+8a2×4π12=22a233+8a2π3=4a233+16a2π3Hence required area is 4a233+16a2π3 sq
units

Find the area of the region{(x,y): y^2 is less than equal to 6 ax and x^2 + y^2 is lessor equal to 16 a^2 using method of integration