Question

the equation of the curve OP is y = x2. find the volume generated when the ares OAP is revolved about a) OX, b) OY, c) AP, d) BP. here's the points: B(0,9), O(0,0), P(3,9), A(3,0)

Answer 1

Right?

Answer 2

Well I don't have time to do all of them but I do have time for one of them is that okay?

Answer 3

A should be in B's psition and B should be in a's

Answer 4

thats ok, thanks for the time :)

Answer 5

Oh, my bad. I get the gist of it though. I'll do a), then i'll go to bed *YAWN lol if I mess up, you know why haha. So here it goes: ready?

Answer 6

tnx! :)

Answer 7

a) wants us to revolve the function \(y=x^2\) about the x axis from \(x\in [0,3]\). In general, the formula for volume generated by revolving of a function \(f(x)\) about the x-axis in the interval \([a,b]\) is defined by:
\[V=\pi\int^b_a{[f(x)]^2dx}\]
So therefore, let's set up the integral:
\[\eqalign{
V&=\pi\int^3_0[x^2]^2dx \\
&=\pi\int^3_0{x^4}dx \\
&=\pi\left.\left(\frac{x^5}{5}\right)\right|^3_0 \\
&=\pi\left(\frac{(3)^5}{5}-\frac{(0)^5}{5}\right) \\
&=\pi\left(\frac{243}{5}\right) \\
&=\frac{243\pi}{5}
}\]

Answer 8

PHEW. Cool?

Answer 9

yeah!! thanks a lot. are u going to bed already?

Answer 10

YEAH? Haha It's 2:33AM and I HAVE A COLD hahaha is there any reason I shouldn't be?

Answer 11

Haha no honestly im really tired. I can finish this with you later though if ya want! Just "fan" me and the next time im online like today at night (like 21 hours from now haha), tomorrow, or the day after, give me a shout!

Answer 12

well, that's okay. u really helped me alot! thnx. have a good sleep :)

Answer 13

Thanks a lot Ella! Ill try to be on like tomorrow at the same time like (11:00PM) which would be like 20 hours from now

Answer 14

okay. have a good sleep :)

Answer 15

help here! :((((((((((

Answer 16

For the revolution around the y axis is something similar. In this case,
\[y=x^2⇒x=\sqrt{y}\]
For the following operations we don't need the plus/minus sign we usually use in square root. Then, following the previous calculus,
\[V=\pi\int_0^9(\sqrt{y})^2dy=81\pi/2\]

Answer 17

For c) and d), they are a variation of the one that me and @John_ES was doing. The concept is that if they need us to revolve it around a line such as \((x=a)\) or \((y=b)\), then we transform the original function so that the line becomes one of the axes! So when they tell us for d) to revolve around the line BP (where, they are telling us to revolve around the line \((y=9)\). So lets transform the original function by moving it 9 units down. The original function is \(f(x)\) and the transformed one will be \(g(x)\):
\[\eqalign{
&f(x)=x^2 \\
&g(x)=x^2-9 \\
}\]
Now we need to revolve g(x) about the x-axis from 0 to 3:
\[\eqalign{
V&=\pi\int^3_0{(x^2-9)^2}\phantom{.}dx \\
&=\pi\int^3_0{x^4-18x^2+81}\phantom{.}dx \\
&=\pi\left.\left(\frac{x^5}{5}-\frac{18x^3}{3}+81x\right)\right|^3_0 \\
&=\pi\left.\left(\frac{x^5}{5}-6x^3+81x\right)\right|^3_0 \\
&=\pi\left.\left(\frac{x^5}{5}-6x^3+81x\right)\right|_{x=3} \\
&=\pi\left(\frac{243}{5}-162+243\right) \\
&=\pi\left(\frac{6(243)}{5}-162\right) \\
&=\pi\left(\frac{1458-810}{5}\right) \\
&=\pi\times\frac{648}{5} \\
&=\frac{648\pi}{5}units
}\]

Answer 18

I'll do c) tomorrow :)

Answer 19

For c), we need to revolve around the vertical line AP which is the linear line \(x=3\). So, similarly to d), we need to move the function \((f(x)=x^2)\) left 3 units to obtain \((g(x)=(x+3)^2)\)
So now we need to revolve the function g(x) about the y-axis. Lets transform into a function of y, \(h(y)\):
\[\eqalign{
&g(x)=(x+3)^2 \\
&y=(x+3)^2 \\
&\pm\sqrt{y}=x+3 \\
&x=-3\pm\sqrt{y} \\
&h(y)=-3\pm\sqrt{y} \\
}\]
So now we need to revolve \(h(y)\) about the y-axis from \(x\in[-3,0]\) which is the interval \(y\in[0,9]\). Visually, we can see that we are actually revolving only the plus side \(h(y)=-3+\sqrt{y}\) So we have the integral:
\[\eqalign{
V&=\pi\int^9_0{(-3+\sqrt{y})^2}\phantom{.}dy \\
&\\
&=\frac{27\pi}{2}units^3\\
}\]