Volumes and Areas (1 Viewer)

[Avenger6]

Hi all, im struggling with finding the volumes and areas in these questions. I have solved part (a) in both of these questions but cannot solve part (b) in both:

I made out 53a) to equal ln (5/2) units^2 and 55a) to equal 1.29 units^3 which are both correct according to the answers. However, I have no idea on how to solve part (b) for both of these questions.

Help is very much appreciated .

 

[vds700]

Hi all, im struggling with finding the volumes and areas in these questions. I have solved part (a) in both of these questions but cannot solve part (b) in both:

I made out 53a) to equal ln (5/2) units^2 and 55a) to equal 1.29 units^3 which are both correct according to the answers. However, I have no idea on how to solve part (b) for both of these questions.

Help is very much appreciated .

53b)
V =
5
pi. ∫(1/x)^2 dx
2
=pi .∫x^-2 dx [5, 2]
=pi . [(-x^-1)] 5, 2
=pi . [-1/x] 5, 2
=pi [(-1/5) + (1/2)]
3pi/10 cubic units

55b) y = lnx
x = e^y
when x = 1 , y = 0, when x = 3, y = ln3
A = ∫e^y dy [ln3, 0]
= [e^y] ln3, 0
= 3 - 1
=2

area of rectangle enclosing the whole area is 2ln3
therefore required area is 2ln3 - 2

 

[Avenger6]

Thanks for the response. Im still a little confused however... In question 53b) i don't see how you got pi . [(-x^-1)] 5, 2 from pi .∫x^-2 dx [5, 2]. And in question 55b) i do not understand the last part about the area of rectangle enclosing the whole area...isn't 2 the area??? And the answer the book gives me for 55b) is 3ln3-2 units^3. Thanks again for the help.

 

[vds700]

Thanks for the response. Im still a little confused however... In question 53b) i don't see how you got pi . [(-x^-1)] 5, 2 from pi .∫x^-2 dx [5, 2]. And in question 55b) i do not understand the last part about the area of rectangle enclosing the whole area...isn't 2 the area??? And the answer the book gives me for 55b) is 3ln3-2 units^3. Thanks again for the help.

sorry mate i should have stepped my working out abit more.

in 53 b, we need to integrate 1/x^2, this can be written as x^-2
to integrate this, we use the standard integral

∫x^n dx = x^(n+1)/(n+1)
so
∫x^-2 dx = X^-1/-1 = -x^-1 =-1/x using negative index laws



in 55b)

see my diagram. The blue area is 2, but we want the area with respect to the x-axis under the curve between 1 and 3.

the area enclosing both areas has an area of 2ln3 (a = b x h)

so to find the area under the curve, you subtract the area with respect to the y-axis from the area of the rectangle.

so the area is 2ln3 - 2, the textbook is slightly wrong, they got the area of the rectangle wrong.

 

[Avenger6]

Of course...that makes much more sense. Thank you for the help.