# 2 questions on area between curves! help please!

• Jun 22nd 2008, 02:41 PM
sgares
2 questions on area between curves! help please!
Ahh, yikes. I just typed out all of the images to prevent any further mishaps hah.

Find c > 0 such that the area of the region enclosed by the parabolas
y = x^2 - c^2 and y = c^2 - x^2 is 30.

Find the area of the region bounded by the parabola y=4x^2, the tangent line to this parabola at (3,36) and the x axis.

Ok, so my method of solving the first one was kind of strange, and I'm fairly certain that I used a completely incorrect method, but here it is...
I assumed that if I found the area in two quadrants then I could just multiply it by 2 and get the total area. So I did y = x^2 - c^2 = area of 15. Then I integrated, plugged in -c and c for x and solved but I came up with a wrong answer, which I think was something like 2.83.

For the second problem I found the began by finding the derivative which was 8x then found the equation for the line by plugging in 36, 3, 8 into y=mx+b equation, which ended up being y=8x+12. Then I integrated
S 8x+12 - 4x^2 and I'm not sure what to even use as the limits at this point, but I made multiple guesses and was wrong on all of them.

• Jun 22nd 2008, 02:42 PM
Mathstud28
Quote:

Originally Posted by sgares
Find https://webwork2.la.asu.edu/webwork2...a0ded42b71.png such that the area of the region enclosed by the parabolas https://webwork2.la.asu.edu/webwork2...9d5ae113c1.png and https://webwork2.la.asu.edu/webwork2...8ece8220d1.png is https://webwork2.la.asu.edu/webwork2...7686503371.png.

Find the area of the region bounded by the parabola https://webwork2.la.asu.edu/webwork2...60acdc32a1.png, the tangent line to this parabola at https://webwork2.la.asu.edu/webwork2...3f6f6377c1.png and the https://webwork2.la.asu.edu/webwork2...dd0b8b8e91.png axis.

Ok, so my method of solving the first one was kind of strange, and I'm fairly certain that I used a completely incorrect method, but here it is...
I assumed that if I found the area in two quadrants then I could just multiply it by 2 and get the total area. So I did https://webwork2.la.asu.edu/webwork2...8ece8220d1.png = area of 15. Then I integrated, plugged in -c and c for x and solved but I came up with a wrong answer, which I think was something like 2.83.

For the second problem I found the began by finding the derivative which was 8x then found the equation for the line by plugging in 36, 3, 8 into y=mx+b equation, which ended up being y=8x+12. Then I integrated
S 8x+12 - 4x^2 and I'm not sure what to even use as the limits at this point, but I made multiple guesses and was wrong on all of them. :(

• Jun 22nd 2008, 02:48 PM
sgares
Ahh, yikes. I just typed out all of the images to prevent any further mishaps hah.

Find c > 0 such that the area of the region enclosed by the parabolas
y = x^2 - c^2 and y = c^2 - x^2 is 30.

Find the area of the region bounded by the parabola y=4x^2, the tangent line to this parabola at (3,36) and the x axis.

Ok, so my method of solving the first one was kind of strange, and I'm fairly certain that I used a completely incorrect method, but here it is...
I assumed that if I found the area in two quadrants then I could just multiply it by 2 and get the total area. So I did y = x^2 - c^2 = area of 15. Then I integrated, plugged in -c and c for x and solved but I came up with a wrong answer, which I think was something like 2.83.

For the second problem I found the began by finding the derivative which was 8x then found the equation for the line by plugging in 36, 3, 8 into y=mx+b equation, which ended up being y=8x+12. Then I integrated
S 8x+12 - 4x^2 and I'm not sure what to even use as the limits at this point, but I made multiple guesses and was wrong on all of them. :(

• Jun 22nd 2008, 02:56 PM
Mathstud28
Quote:

Originally Posted by sgares
Ahh, yikes. I just typed out all of the images to prevent any further mishaps hah.

Find c > 0 such that the area of the region enclosed by the parabolas
y = x^2 - c^2 and y = c^2 - x^2 is 30.

Find the area of the region bounded by the parabola y=4x^2, the tangent line to this parabola at (3,36) and the x axis.

Ok, so my method of solving the first one was kind of strange, and I'm fairly certain that I used a completely incorrect method, but here it is...
I assumed that if I found the area in two quadrants then I could just multiply it by 2 and get the total area. So I did y = x^2 - c^2 = area of 15. Then I integrated, plugged in -c and c for x and solved but I came up with a wrong answer, which I think was something like 2.83.

For the second problem I found the began by finding the derivative which was 8x then found the equation for the line by plugging in 36, 3, 8 into y=mx+b equation, which ended up being y=8x+12. Then I integrated
S 8x+12 - 4x^2 and I'm not sure what to even use as the limits at this point, but I made multiple guesses and was wrong on all of them. :(

The first one is formally unanswerable, but informally

Let $c^2-x^2

Now then you must find c by solving the following(Easy) integral equation

$\int_a^{b}\bigg[(x^2-c^2)-(c^2-x^2)\bigg]dx=30$

The tangent line is given by

$y-36=24(x-3)$

Now let $g(x)=24(x-3)+36$

and $f(x)=4x^2$

Now find the two values of x such that

$f(x)=g(x)$

Then test one value in each to show that either

$\forall{x}\in(a,b)\quad{f(x)}

Then the area would be

$A=\int_a^{b}g(x)-f(x)dx$

or

$A=\int_a^{b}f(x)-g(x)dx$

respectively

Good luck, and if you have any more questions, dont hesitate to ask.

Mathstud
• Jun 22nd 2008, 03:15 PM
sgares
Thank you for the help, I really do appreciate it! I'll get back to working on them now! =)