# Tutor profile: Vinay S.

## Questions

### Subject: Calculus

Find the volume formed by rotating the region enclosed by: x=6y and y^3=x with y>=0 about y-axis.

First find points of intersection. 6y=y^3 y^3-6y=0 y(y^2-6)=0 y=0 or y^2-6=0 y=0 or y=√6 ( Did not include negative value as y>=0) Now we use washer’s method: And so we get V =π∫((6y)^2-(y^3 )^2 )dy {integrate from 0 to√6) = π∫(36y^2-y^6 )dy = π|12y^3-y^7/7|{limits 0 and √6) Now plug y=√6 and simplify. So V = π(72√6-(216√6)/7=(288√6 π)/7=100.78π

### Subject: Calculus

∫x√(2x^2+6)

Here we need to use u-substitution. Use u=2x^2+6 So du = 4xdx or dx=du/4x So new integral after plugging in values for (2x^2+6) and dx would be: ∫(√u du)/4 1/4 ∫(√u du) Now we can write it as : 1/4 ∫(u^(1/2) du) Use rules of integral to integrate this. Then plug back u = 2x^2+6.

### Subject: Algebra

Dave gets a commission of $300 on every car he sells. He is expected to sell a minimum of 10 cars a month. If he exceeds this by more than 20%, he is paid an extra $75 commission per car on all the cars sold during the month.

Answer: 5. If Dave sells 12 cars he will get a commission of 12($300)=$3,600. It means Dave must have sold more than 12 cars as the commission earned is more than $3,600. Commission per car now becomes = $375. If the number of cars he sold this month was "a", then 375a=5,625. On solving, we get a = 15, which is 5 more than 10.

## Contact tutor

needs and Vinay will reply soon.