Integration - Cambridge (1 Viewer)

XShiv · Jan 12, 2010

Hi guys,
i have just got a question from the cambridge year 11 3-unit book on volumes that is making me pull my hair out, i think i'm getting my limits wronge but anyways it's"

Excercise 11G Question 19 Part (C)

i have got parts (a) and (b) correct i think...
Click for my graph

Thanks

cutemouse · Jan 12, 2010

Post the question.

XShiv · Jan 12, 2010

Posted.

Drongoski · Jan 12, 2010

Q19 c

$\100dpi y = f(x)\ = x^{3}+2\\ \\ f'(x) = 3x^{2}\\ \\ f(1) = 3\ \ \ \ f'(1) = 3\\ \\ \therefore \ tangent\ @\ A(1,3)\ \ is:\ \ y = 3x\\ \\ i)\ \ \ vol \ about\ x-axis\\ \\ V_{x}\ =\ \pi \int_{0}^{1}(y_{2}^{2}-y_{1}^{2})\ dx\\ \\ \ =\ \ \pi \int_{0}^{1}((x^{3}+2)^{2}\ -\ (3x)^{2})\ dx\ \ =\ \ \pi \int_{0}^{1}[x^{6}+4x^{3}+4-9x^{2}]\ dx\\ \\ =\ \ \pi \left [ \frac{x^{7}}{7} +x^{4}-3x^{3}+4x\right ]^{1}_{0}\ \ =\ \ \frac{15}{7}\pi \ \ unit^{3}\\ \\ \\ ii)\ \ \ vol\ about\ \ y-axis\\ \\ V_{y}\ =\ \int_{0}^{1}2\pi x(y_{2}-y_{1})\ dx\ \ =\ \ 2\pi \int_{0}^{1}x(x^{3}+2-3x)\ dx\\ \\=\ \ 2\pi \left [ \left \frac{x^{5}}{5}-x^{3} +x^{2}\right ]^{1}_{0}\ \ =\ \ \frac{2\pi }{5}\ \ unit^{3}$

XShiv · Jan 12, 2010

Thanks alot, appreciate it =]

davidgoes4wce · Feb 14, 2016

Drongoski said:
Q19 c

$\100dpi y = f(x)\ = x^{3}+2\\ \\ f'(x) = 3x^{2}\\ \\ f(1) = 3\ \ \ \ f'(1) = 3\\ \\ \therefore \ tangent\ @\ A(1,3)\ \ is:\ \ y = 3x\\ \\ i)\ \ \ vol \ about\ x-axis\\ \\ V_{x}\ =\ \pi \int_{0}^{1}(y_{2}^{2}-y_{1}^{2})\ dx\\ \\ \ =\ \ \pi \int_{0}^{1}((x^{3}+2)^{2}\ -\ (3x)^{2})\ dx\ \ =\ \ \pi \int_{0}^{1}[x^{6}+4x^{3}+4-9x^{2}]\ dx\\ \\ =\ \ \pi \left [ \frac{x^{7}}{7} +x^{4}-3x^{3}+4x\right ]^{1}_{0}\ \ =\ \ \frac{15}{7}\pi \ \ unit^{3}\\ \\ \\ ii)\ \ \ vol\ about\ \ y-axis\\ \\ V_{y}\ =\ \int_{0}^{1}2\pi x(y_{2}-y_{1})\ dx\ \ =\ \ 2\pi \int_{0}^{1}x(x^{3}+2-3x)\ dx\\ \\=\ \ 2\pi \left [ \left \frac{x^{5}}{5}-x^{3} +x^{2}\right ]^{1}_{0}\ \ =\ \ \frac{2\pi }{5}\ \ unit^{3}$

I couldn't see your working so I decided to redo it again:

Here is the graph:
Q19b.

19 c (i) and (ii)

and

KingOfActing · Feb 14, 2016

davidgoes4wce said:
I couldn't see your working so I decided to redo it again:
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There's no reason to bump a 6 year old thread...

davidgoes4wce · Feb 14, 2016

KingOfActing said:
There's no reason to bump a 6 year old thread...

It's just some good practice for the current HSC students to go through again.

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Integration - Cambridge (1 Viewer)

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