Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
2.3k views
in Definite Integrals by (29.3k points)
closed by

Area enclosed between the curve y2 (2a – x) = x3 and the line x = 2a above x-axis is

A. \(\pi a^2\)

B. \(\frac{3}{2}\pi a^2\)

C. 2π a2

D. 3π a2

1 Answer

+2 votes
by (28.8k points)
selected by
 
Best answer

Correct answer is B.

Since we are concerned with the area above the x – axis, we’ll be considering the positive root.

We can see that at x = 0, y = 0.

So,

Putting √x = u

We get du = (1/2√x)dx

or, dx = 2u du

So,

Putting u = √2√a sin t

We get du = √2√a cos t dt

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...