Volumes of revolution question

Figure 6 shows part of an ellipse with equation x^2/16 + (y-10)^2/64 = 1 inside a cylinder with a diamater of 10 cm and a height of 10 cm. The shaded region is rotated 360° about the y-axis to generate a solid of revolution. Find the exact volume of solid generated. Leave your answer in the form an, where a is a rational number to be found.

https://media.cheggcdn.com/study/999/999a2c4a-9e67-481d-b870-e47f2a03106a/image

I believe the answer I got to is 494/3pi by doing the volume of the cylinder pi*r^2*h - the integral between 10 and 2 of pi*x^2 dy.
However Answers on GauthMaths say that it is 590pi.
Did I do anything wrong?

Reply 1

mqb2766

Original post by sgdumpling

Figure 6 shows part of an ellipse with equation x^2/16 + (y-10)^2/64 = 1 inside a cylinder with a diamater of 10 cm and a height of 10 cm. The shaded region is rotated 360° about the y-axis to generate a solid of revolution. Find the exact volume of solid generated. Leave your answer in the form an, where a is a rational number to be found.
https://media.cheggcdn.com/study/999/999a2c4a-9e67-481d-b870-e47f2a03106a/image
I believe the answer I got to is 494/3pi by doing the volume of the cylinder pi*r^2*h - the integral between 10 and 2 of pi*x^2 dy.
However Answers on GauthMaths say that it is 590pi.
Did I do anything wrong?

The cylinder volume is 250pi so it has to be less than that (so 590pi must be wrong), but upload your working (an image) if you want it checked. It sounds like the right approach and the value sounds about right.

(edited 11 months ago)

Reply 2

DFranklin

Original post by sgdumpling

Figure 6 shows part of an ellipse with equation x^2/16 + (y-10)^2/64 = 1 inside a cylinder with a diamater of 10 cm and a height of 10 cm. The shaded region is rotated 360° about the y-axis to generate a solid of revolution. Find the exact volume of solid generated. Leave your answer in the form an, where a is a rational number to be found.
https://media.cheggcdn.com/study/999/999a2c4a-9e67-481d-b870-e47f2a03106a/image
I believe the answer I got to is 494/3pi by doing the volume of the cylinder pi*r^2*h - the integral between 10 and 2 of pi*x^2 dy.
However Answers on GauthMaths say that it is 590pi.
Did I do anything wrong?

I agree with your answer.

[if you take a half-sphere of radius 4 and stretch it along the y-axis by a factor of 2 you get the ellipsoid. so the ellipsoid has volume

\frac{4}{3} \pi 4^3 \times \frac{1}{2} \textrm{(for the half-sphere)} \times 2 \textrm{(for the y-scaling)} = 4^4 \pi / 3

. Then subtract from the cylinder volume.]

(edited 11 months ago)

Reply 3

SGDumpling

Original post by mqb2766

This is my working!

Thank you in advanced

Reply 4

mqb2766

Looks good. You could have simplified it a touch as
x^2 = 16 - (y-10)^2/4
so youre integrating that from 2 to 10. Doing a simple sub of u = (y-10) means youre integrating
16 - u^2/4
from -8 to 0.

Reply 5

SGDumpling

Original post by mqb2766

Looks good. You could have simplified it a touch as
x^2 = 16 - (y-10)^2/4
so youre integrating that from 2 to 10. Doing a simple sub of u = (y-10) means youre integrating
16 - u^2/4
from -8 to 0.

Thank you so much! We havent learnt u substitution yet in school, but it seems very useful, i could try look into it before school like most things anyways.

Reply 6

SGDumpling

Original post by DFranklin

I agree with your answer.
[if you take a half-sphere of radius 4 and stretch it along the y-axis by a factor of 2 you get the ellipsoid. so the ellipsoid has volume

\frac{4}{3} \pi 4^3 \times \frac{1}{2} \textrm{(for the half-sphere)} \times 2 \textrm{(for the y-scaling)} = 4^4 \pi / 3

. Then subtract from the cylinder volume.]

Thank you very much!

Reply 7

mqb2766

Original post by sgdumpling

Thank you so much! We havent learnt u substitution yet in school, but it seems very useful, i could try look into it before school like most things anyways.

I think you do substitution in year 2, so youre probably right. You could simply note the integral of
(y-10)^2
is
(y-10)^3/3
(plus constant) which is easy enough to verify by differentiating the result. So you dont really need substitution (or to expand the quadratic in order to integrate it)