Question

Hi! Need help regarding calculus.

The region bounded by the curve y=x^2 +1 and the line y=x+3. Find the area of the region using horizontal strip. Find also the volume of the region if the axis of revolution is y=8.

Thanks!

Answer 1

www.most.gov.mm/techuni/media/EM_02011_chap456.pdf

Answer 2

@rohangrr, please let people know what the link is you are posting if it is a huge pdf document (41 pages seems excessive)

Answer 3

\[A = \int\limits_{-1}^{2}(x+3)-(x^{2}+1) dx\]

Answer 4

For volume, you are summing up the areas of the cross-sections
each cross-section is a ring with inner radius and outer radius
inner = 8-(x+3) = 5-x
outer = 8-(x^2+1) = 7-x^2
\[V =\pi \int\limits_{-1}^{2}(7-x^{2})^{2}-(5-x)^{2} dx\]

Answer 5

@dumbcow Thanks for the fast reply. But isn't that for the vertical strip for the area? .

Answer 6

oh yeah, sorry didn't think about that but since the functions are in terms of x its easier to integrate wrt x

Answer 7

I was able to compute that for the vertical strip. I think the answer was 9/2. But for some reason I'm having a hard time for the horizontal strip. Must have 2 regions, I guess. Unfortunately, no luck, I'm having diff. answer with horizontal strip.

Answer 8

well because its a parabola, it makes it difficult
for region: 2<y<5
\[A = \int\limits_{2}^{5}\sqrt{y-1} - (y-3) dy\]
for region: 1<y<2
\[A = 2\int\limits_{1}^{2}\sqrt{y-1} dy\]

Answer 9

Got it! Thanks