Find the volume of the solid that results when the region enclosed by the given curves is revolved....?

Question

Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis.

y = 7sqrt(cosx), x=pi/4, x=pi/2, y=0

give answer in exact form.
Thanks!! =D

Answer 1

Model this a a series of cylinders next to each other. The volume of each cylinder is piy^2 times delta x, because y is the radius of each cylinder and the height of each cylinder is delta x. When you add up all the indiviual cylinders, you get the total volume. In calculuse, this is done by integration.

Integrate over dx with the limits x = pi/4 and x=pi/2.

volume = integral of 49*pi*cos(x) dx.

Integral of cos(x) is sin(x), so answer is:

49pi*(sin(pi/2) - sin(pi/4)) = 49pi*(1-sqrt(2)/2)

Answer 2

This is not too hard. If you draw the diagram you'll see that it's a nice easy one, y starts from 7 at x = 0 to 7.2^(-1/4) at x = pi/4 and goes down to 0 at x = pi/2.
So we just need V = pi. int(pi/4 to pi/2) y^2 dx
= pi. int(pi/4 to pi/2) 49 cos x dx
= pi. 49 sin x [pi/4 to pi/2]
= 49 pi [sin pi/2 - sin pi/4]
= 49 pi (1 - 1/sqrt(2))
If you want to rationalise the denominator you can rewrite this as
= 49 pi (1 - sqrt(2) / 2).

Answer 3

First set up the integral,
lower bound pi/4
upper bound pi/2
The integral is pi(7(sqrt(cosx)))^2dx
this reduces to 49pi (outside the integral) and cosx inside the integral
the integral of cosx is sinx
Evaluated it becomes 49pi-[49(pi)sqrt(x)]/2

Answer 4

that would be integral pi * (f(x))^2 =
7 pi integral cos^2 x dx on the interval

Answer 5

locate the quantity of the coolest that consequences at the same time as the area enclosed by using the given curves is revolved about the x-axis. x=?y x=y/4 please remedy this question i want it urgent