Simple Definite Integral: Find the Area Bounded by a Curve

[Mentallic]

Homework Statement

Find the area bounded by the curve y=x^2-2, the y-axis, y=0 and y=1

I got an answer, but when I checked it using graphmatica, it was wrong (also the result is logically too large). I cannot see where I went wrong though, so could someone please help me spot the mistake.

Homework Equations

\int_{a}^{b}f(y)dy = \left [ F(y) \right ]_{a}^{b}

\int_{a}^{b}(ax+b)^ndx=\left [ \frac{(ax+b)^{n+1}}{a(n+1)}\right ]_a^b

The Attempt at a Solution

A=\int_{0}^{1}xdy

where x=\pm \sqrt{y+2}

thus,
A=\int_{0}^{1}\pm \sqrt{y+2}.dy

A=2\int_{0}^{1}\sqrt{y+2}.dy

=2 \left[ \frac{(y+2)^{3/2}}{3/2} \right]_0^1

=2\left( \frac{3^{3/2}}{3/2} \right)

=4\sqrt{3}.u^2

 

[lanedance]

hmm... i think it could be the factor of 2, the question says from y axis, so try only taking +sqrt

 

[Mentallic]

Uh no actually, there is something else going on here.

This is why:

Using graphmatica, I found \int_0^1 \sqrt{x+2}.dx \approx 1.5785

I changed the function to integrate because the program seems to only be able to integrate with respect to the x-axis. The function is still equivalent to the original question though (for positive values of y).

So if I take the area of \int_0^1 \sqrt{y+2}.dy in the positive quadrant only, then I get 2\sqrt{3} \approx 3.464 which is far off the value given from graphmatica (and this answer doesn't seem correct either, from an elementary view of the approx area).

The problem is somewhere else, and I can't seem to spot it...

 

[Mentallic]

Ahh I found where I went wrong:

\left[ \frac{(y+2)^{3/2}}{3/2} \right]^1_0 \neq 2\sqrt{3}

I assumed substituting 0 into the primitive returned 0. I'll be sure not to make that mistake again..

 

[HallsofIvy]

Homework Statement

Find the area bounded by the curve y=x^2-2, the y-axis, y=0 and y=1

I got an answer, but when I checked it using graphmatica, it was wrong (also the result is logically too large). I cannot see where I went wrong though, so could someone please help me spot the mistake.

Homework Equations

\int_{a}^{b}f(y)dy = \left [ F(y) \right ]_{a}^{b}

\int_{a}^{b}(ax+b)^ndx=\left [ \frac{(ax+b)^{n+1}}{a(n+1)}\right ]_a^b

The Attempt at a Solution

A=\int_{0}^{1}xdy

where x=\pm \sqrt{y+2}

That's makes no sense. You must integrate a function and "\pm\sqrt{y+2}" is not a function. The x distance from one side of the graph to the other is 2\sqrt{y+2}. Further, \int_0^1 xdx[/itex] is NOT the same as \int_0^1\sqrt{y+ 2}dy. If you let x= \sqrt{y+2}= (y+ 2)^{1/2}, then dx= (1/2)(y+2)^{-1/2}dy= 1/(2x)dy so dy= 2xdx. Also when y= 0, x= \sqrt{2} and when y= 1, x= \sqrt{3}.<br /> \int_0^1 \sqrt{y+2}dy= 2\int_{\sqrt{2}}^{\sqrt{3}} x^2dx= \frac{2}{3}\left(\sqrt{27}- \sqrt{8}\right) so your area is (4/3)\left(\sqrt{8}- \sqrt{2}\right).<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> thus, <br /> A=\int_{0}^{1}\pm \sqrt{y+2}.dy<br /> <br /> A=2\int_{0}^{1}\sqrt{y+2}.dy<br /> <br /> =2 \left[ \frac{(y+2)^{3/2}}{3/2} \right]_0^1<br /> <br /> =2\left( \frac{3^{3/2}}{3/2} \right)<br /> <br /> =4\sqrt{3}.u^2 </div> </div> </blockquote>

 

[Mentallic]

That's makes no sense. You must integrate a function and "\pm\sqrt{y+2}" is not a function.

How so? if y=x^2-2 is a function, then if x is made the subject, why isn't it a function anymore?

The x distance from one side of the graph to the other is 2\sqrt{y+2}. Further, \int_0^1 xdx[/itex] is NOT the same as \int_0^1\sqrt{y+ 2}dy. If you let x= \sqrt{y+2}= (y+ 2)^{1/2}, then dx= (1/2)(y+2)^{-1/2}dy= 1/(2x)dy so dy= 2xdx. Also when y= 0, x= \sqrt{2} and when y= 1, x= \sqrt{3}.

<br /> <br /> Yes I did this, but you could argue I made it in a less conventional (cheating) way <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /><br /> <br /> <blockquote data-attributes="" data-quote="HallsofIvy" data-source="post: 2093184" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> HallsofIvy said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> \int_0^1 \sqrt{y+2}dy= 2\int_{\sqrt{2}}^{\sqrt{3}} x^2dx= \frac{2}{3}\left(\sqrt{27}- \sqrt{8}\right) so your area is (4/3)\left(\sqrt{8}- \sqrt{2}\right). </div> </div> </blockquote>The area you found was correct but you simplified it incorrectly.<br /> <br /> \frac{2}{3}\left(\sqrt{27}- \sqrt{8}\right) \neq (4/3)\left(\sqrt{8}- \sqrt{2}\right)<br /> <br /> The answer A=\frac{4}{3}(3\sqrt{3}-2\sqrt{2}) \approx 3.157 is also what Graphmatica gave me. Now the question as to whether the area is actually half of this or not. The question says it is also bounded by the y-axis, but y=x^2-2 has x values of both negative and positive between 0&amp;lt;y&amp;lt;1 to the y-axis...