Volume of the Revolution of a Solid (1 Viewer)

SunnyScience · Jan 29, 2012

Don't know how to go these questions :/ - anybody care to help me out please

? (*cough* Spiralflex *cough*)

1. A petrol tank is designed by the rotation of the curve y=1/5x(x-40) about the x-axis between the planes x=0 and x=40. If units are in centrimeters, how many litres would the tank hold?

2. Find the equation of the tangent to the parabola y=2x^2 at (1,2). Calculate its point of intersection with the x-axis and the volume of the solid formed when the area between the parabola, the tangent line and the x-axis is revolved about the x-axis.

And if you have time, it would be nice if you could do these easier ones as well please

3. Find the area bounded by the curve y=(9-x)^0.5 [square root of] and the x and y axes.
find the volume of the solid generated when this area is rotated about:
(i) the x axis
(ii) the y-axis

4. The curve x^2 = 16-4y^2 is rotated about the y-axis. Find the volume of the generated solid.

Sorry for the poor formatting guys - your imput is greatly appreciated

SpiralFlex · Jan 29, 2012

Always draw a pretty diagram (pending)

$y=\pi\int_{0}^{40} [\frac{1}{5}x(x-40)]^2\; dx$

$=\frac{\pi}{25}\int_{0}^{40} x^2(x-40)^2\; dx$

$=\frac{\pi}{25}\int_{0}^{40} x^2[x^2-80x+1600]\; dx$

$=\frac{\pi}{25}\int_{0}^{40}[x^4-80x^3+1600x^2]\; dx$

$=\frac{\pi}{25}[\frac{x^5}{5}-20x^4+\frac{1600}{3}x^3]$ (Sub your limits.)

Note: $1 $cm$^3 = 0.001\; $litres$$

Editing silly mistakes.

We first must find this tangent.

$y=2x^2$

$y'=4x$

At $x=1$

$y'=4$

$y-2=4(x-1)$

$4x-y-2=0$

To find the intersection with the x axis,

Let $y=0$

$x=\frac{1}{2}$

$\therefore (\frac{1}{2},0)$ is the point of intersection.

We need to find the volume enclosed by the curve with the limits 2,0 minus the volume enclosed by the shape that makes a cone.

$V=\pi\int_{0}^{2}4x^4 \; dx-$(Volume of cone) $$

$=\pi[\frac{4}{5}(2)^5]-\frac{1}{3}\pi(2)^2*\frac{1}{2}$

$=...$

3.

$\int_{0}^{9}(9-x)^{\frac{1}{2}}\; dx$

$=[-\frac{2}{3}(9-x)^{\frac{3}{2}}]\; dx$ (With limits)

Substitute it in.

$$About x axis:$\; V=\pi \int_{0}^{9}(9-x)\; dx$

$V=\pi[9x-\frac{x^2}{2}]$ (Sub limits.)

$About y axis: Solve for first.$

$x=9-y^2$

$x^2=-y^4-18y^2+81$

$V=\pi \int_{0}^{9}(-y^4-18y^2+81)\; dy$ (Sub your limits.)

$4. This curve is an ellipse. Stretched horizontally by a factor of 4 and vertically by a factor of 2.$

$x^2 = 16-4y^2$

$V=2\pi\int_{0}^{2}(16-4y^2)\; dy$

$=2\pi[16y-\frac{4}{3}y^3]$ (Limits, I am sure you can do this.)

SunnyScience · Jan 29, 2012

First off, OMFG spiral, this is simply amazing - you have no idea how gob smacked i am by your answers - jesus - thankyou!

However, (i must be doing something wrong)

When i sub in the limits, the answers come out wrong

?

Can u please help me with why this is?

pi/25[x^5/5 - 40x^4 + 1600/3 x^3] 40/0 (limits)

= pi/25[40^5/5 - 40(40^4) + 1600/3 (40)^3] - (0)
= pi/25[-4778666.67]
= -6005049.638

But the answer is 428.932L? - Help? :/

SpiralFlex · Jan 29, 2012

20x^4 not 40x^4 I was editing it. I need to watch tennis now! bye!

D94 · Jan 29, 2012

SunnyScience said:
First off, OMFG spiral, this is simply amazing - you have no idea how gob smacked i am by your answers - jesus - thankyou!

However, (i must be doing something wrong)

When i sub in the limits, the answers come out wrong ?

Can u please help me with why this is?

pi/25[x^5/5 - 40x^4 + 1600/3 x^3] 40/0 (limits)

= pi/25[40^5/5 - 40(40^4) + 1600/3 (40)^3] - (0)
= pi/25[-4778666.67]
= -6005049.638

But the answer is 428.932L? - Help? :/

Should be 20x^4 (not 40)

edit: spiral beat me

bleakarcher · Jan 29, 2012

SpiralFlex said:
Always draw a pretty diagram (pending)

$y=\pi\int_{0}^{40} [\frac{1}{5}x(x-40)]^2\; dx$

$=\frac{\pi}{25}\int_{0}^{40} x^2(x-40)^2\; dx$

$=\frac{\pi}{25}\int_{0}^{40} x^2[x^2-80x+1600]\; dx$

$=\frac{\pi}{25}\int_{0}^{40}[x^4-80x^3+1600x^2]\; dx$

$=\frac{\pi}{25}[\frac{x^5}{5}-20x^4+\frac{1600}{3}x^3]$ (Sub your limits.)

Note: $1 $cm$^3 = 0.001\; $litres$$

Editing silly mistakes.

We first must find this tangent.

$y=2x^2$

$y'=4x$

At $x=1$

$y'=4$

$y-2=4(x-1)$

$4x-y-2=0$

To find the intersection with the x axis,

Let $y=0$

$x=\frac{1}{2}$

$\therefore (\frac{1}{2},0)$ is the point of intersection.

We need to find the volume enclosed by the curve with the limits 2,0 minus the volume enclosed by the shape that makes a cone.

$V=\pi\int_{0}^{2}4x^4 \; dx-$(Volume of cone) $$

$=\pi[\frac{4}{5}(2)^5]-\frac{1}{3}\pi(2)^2*\frac{1}{2}$

$=...$

3.

$\int_{0}^{9}(9-x)^{\frac{1}{2}}\; dx$

$=[-\frac{2}{3}(9-x)^{\frac{3}{2}}]\; dx$ (With limits)

Substitute it in.

$$About x axis:$\; V=\pi \int_{0}^{9}(9-x)\; dx$

$V=\pi[9x-\frac{x^2}{2}]$ (Sub limits.)

$About y axis: Solve for first.$

$x=9-y^2$

$x^2=-y^4-18y^2+81$

$V=\pi \int_{0}^{9}(-y^4-18y^2+81)\; dy$ (Sub your limits.)

$4. This curve is an ellipse. Stretched horizontally by a factor of 4 and vertically by a factor of 2.$

$x^2 = 16-4y^2$

$V=2\pi\int_{0}^{2}(16-4y^2)\; dy$

$=2\pi[16y-\frac{4}{3}y^3]$ (Limits, I am sure you can do this.)

Spiralflex, too much mistakes lately dude, in question two you had to have the first integral from x=0 to x=1 not to x=2.

SpiralFlex · Jan 29, 2012

bleakarcher said:
Spiralflex, too much mistakes lately dude, in question two you had to have the first integral from x=0 to x=1 not to x=2.

Yeah haha I make alot of mistakes.

I did this question in the toilet. But OP should be able to pick that up. I was merely telling him what to do.

Carrotsticks · Jan 29, 2012

SpiralFlex said:
I did this question in the toilet.

Oh and goodbye to 3-digit post numbers.

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Volume of the Revolution of a Solid (1 Viewer)

SunnyScience

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