The diagram shows the curve y =sqrt(3x + 1) and the points P (0, 1) and Q(1, 2) on the curve. The shaded region is bounded by the curve, the y-axis and the line y = 2.
(i) Find the volume obtained when the shaded region is rotated through 360degress about the x-axis.

Tangents are drawn to the curve at the points P and Q.
(ii) Find the acute angle, in degrees correct to 1 decimal place, between the two tangents.

Diagram: http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w08_qp_1.pdf

Question

The diagram shows the curve y =sqrt(3x + 1) and the points P (0, 1) and Q(1, 2) on the curve. The shaded region is bounded by the curve, the y-axis and the line y = 2.
(i) Find the volume obtained when the shaded region is rotated through 360degress about the x-axis.

Tangents are drawn to the curve at the points P and Q.
(ii) Find the acute angle, in degrees correct to 1 decimal place, between the two tangents.

Diagram: http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w08_qp_1.pdf

lxelle · Answer

As for i) 
V= 5/2pi

gorv · Answer

it will be  integration  in 3d space

lxelle · Answer

I supposed I've got to minus something else to find the shaded region.

lxelle · Answer

@gorv Yes, I know it's an integration solving ques.

amistre64 · Answer

do you know how do do a disc method?

amistre64 · Answer

find the volume of the 'outside' shape, then subtract from the the volume of the 'inside' shape

aum · Answer

i) $$
\int_{0}^1\pi (2^2 - y^2)dx = \int_{0}^1\pi (4 - 3x-1)dx = \pi\int_{0}^1 (3 - 3x)dx = 
3\pi \int_{0}^1 (1 - x)dx = ?
$$

lxelle · Answer

The 2^2 part. Where did you get it from?

aum · Answer

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