Help with calculus area problem please :)?
- 2 months ago
You may find this problem easier if you tilt your head 90 degrees: we're just going to integrate on the y-axis.
Integrate f(y) = y/sqrt(49-y^2) from 0 to 5.
First, let's find the indefinite intregral. We have a function of y^2 in the denominator, and a y on the outside, so let's use substitution.
u = 49 - y^2, du = -2y dy, y dy = -1/2 du
So ∫ y dy/sqrt(49-y^2) = ∫ (-1/2 du)/sqrt(u) = -1/2∫u^-1/2 du
Using the power rule: ∫u^-1/2 du = 2u^1/2 + C
Multiplying by the -1/2 on the outside we have -u^1/2 + C, and by reverse-substituting we get that the indefinite integral is -sqrt(49 - y^2).
Now, evaluate that at 0 and at 5 to get the definite integral.
-sqrt(49 - 5^2) - (-sqrt(49)) = -4.8990 + 7 = 2.101 (to 3 d.p.)
Hope this helped!