• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Volumes: Using 3d objects as slice question? (1 Viewer)

kayven

Member
Joined
Jul 8, 2011
Messages
69
Gender
Undisclosed
HSC
2012
Hey guys,

Can you please solve this question for me?
It goes on the lines of: The base of a circle has the equation x^2 + y^2 = 9. Find the volume of the solid if every cross section perpendicular to the y-axis is a square with one side in the base of the circle.


Thankyou so much!
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
You will need to draw 2 diagrams.

1. The circle with radius 3, centred at the origin. Shade a thin vertical slice (or horizontal, but I like vertical because horizontal looks fat) which has thickness delta(x) and its length/height would be 2y.

2. A square which would have side length 2y, as above.

The area of the slice is (2y)(2y) = 4y^2. However, it has thickness delta(x) so the volume of our slice is 4y^2 * delta(x).

Now, we integrate the expression for our slice from x=-3 to x=3 so our integral becomes:



But we have to integrate with respect to X because we have dx. Our expression is in terms of y, so we have to change it back in terms of x by perhaps using the equation of the circle x^2+y^2=9.

So our integral now is:



Then evaluate it as usual.
 

kayven

Member
Joined
Jul 8, 2011
Messages
69
Gender
Undisclosed
HSC
2012
Thanks Carrotsticks, but shouldnt the thickness be delta y considering that the slice is taken perpendicular to the y=axis?
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Thanks Carrotsticks, but shouldnt the thickness be delta y considering that the slice is taken perpendicular to the y=axis?
Not quite. But actually, the question can be done either way. Your slice would even be in the form y=mx+b because of the rotational symmetry of the circle! But of course that would make things unnecessarily complicated.

 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top