# [SOLVED]integration of area between curve and y-axis: transposition question/problem

#### DeusAbscondus

##### Active member
Find the area under curve $y=243x^5$ and between y=1 and y=32

Here is my working out:
1. transpose to make x the subject

$$x=\frac{y^{1/5}}{3}$$

2. integrate in y

$$\int^{32}_1 \frac{y^{1/5}}{3}\cdot dy=(\frac{5\cdot 32^{6/5}}{18})-(\frac{5}{18})=17.5$$

Which is discrepant with given solution of 157.5

Question 1:
Would someone kindly check my answer as I'm beginning to suspect that I have been detained for hours by an error in the solution (ie: $157.5$ could be $17.5$ with a careless interpolation of $5$ between first 2 digits of my answer)?

Question 2: How could one have rewritten the original function so as to translate its curve by 90 degrees and integrate along the x-axis?

Deus Abs

Last edited:

#### Chris L T521

##### Well-known member
Staff member
Re: integration of area between curve and y-axis: transpostion question/problem

Find the area under curve $y=243x^2$ and between y=1 and y=32

Here is my working out:
1. transpose to make x the subject

$$x=\frac{y^{1/5}}{3}$$

2. integrate in y

$$\int^{32}_1 \frac{y^{1/5}}{3}\cdot dy=(\frac{5\cdot 32^{6/5}}{18})-(\frac{5}{18})=17.5$$

Which is discrepant with given solution of 157.5

Question 1:
Would someone kindly check my answer as I'm beginning to suspect that I have been detained for hours by an error in the solution (ie: $157.5$ could be $17.5$ with a careless interpolation of $5$ between first 2 digits of my answer)?

Question 2: How could one have rewritten the original function so as to translate its curve by 90 degrees and integrate along the x-axis?

Deus Abs
I think you meant to say $y=243x^5$ if you ended up with $x=\frac{1}{3}y^{1/5}$.

Now, the area the integral you came up does not represent the area you're supposed to be finding! Keep in mind that $x=2$ when $y=32$ and $x=1$ when $y=1$. The original problem asked you to find the area beneath the curve $y=243x^5$ and between $y=1$ and $y=32$; thus if you're going to rewrite the function in the form $x=f(y)$, then you'll have to find the area "above" that curve (above in the sense of to the right of the function in your standard xy coordinate system). This is where we use the fact that $x$ can not exceed 2. Thus, the function you're supposed to be integrating is $2-\frac{1}{3}y^{1/5}$; the area integral should be
$\int_1^{32}\left(2-\frac{1}{3}y^{1/5}\right)\,dy.$

I hope this makes sense!

#### DeusAbscondus

##### Active member
Re: integration of area between curve and y-axis: transpostion question/problem

$$(\frac{5\cdot 32^{6/5}}{18})=(\frac{5\cdot 32\cdot 2}{18})=\frac{320}{18}$$
Doesn't it?

If so, then $$\frac{320}{18}-\frac{5}{18}=17.5 \ area\ units$$

#### Amer

##### Active member
Re: integration of area between curve and y-axis: transpostion question/problem

Hey
for question like this I like to draw the curve
$$y = 243x^{5}$$
$$y = 1$$
$$y = 32$$ I find the point of intersection between y=1 and $$y = 243x^5$$

$$1 = 243 x^5 \Rightarrow x = \sqrt{\frac{1}{243}} = \frac{1}{3}$$

and between y =32 with $$y = 243x^5$$

$$32 = 243 x^5 \Rightarrow x = \sqrt{\frac{32}{243}} = \frac{2}{3}$$

and the area with the white color
$$A = \int_{\frac{1}{3}}^{\frac{2}{3}} 243x^5 - 1 dx = \frac{243x^6}{6} - x \mid_{\frac{1}{3}}^{\frac{2}{3}}=$$$$\frac{243(2)^6}{3^6(6)} - \frac{2}{3} - \left( \frac{243}{3^6(6)} - \frac{1}{3} \right) = \\ \frac{64}{18} - \frac{2}{3} - \left( \frac{1}{18} - \frac{1}{3}\right)\\ = \frac{63}{18} - \frac{6}{18} = \frac{57}{18}$$

but my answer is not like yours lol

#### earboth

##### Active member
Re: integration of area between curve and y-axis: transpostion question/problem

Find the area under curve $y=243x^5$ and between y=1 and y=32

Here is my working out:
1. transpose to make x the subject

$$x=\frac{y^{1/5}}{3}$$

2. integrate in y

$$\int^{32}_1 \frac{y^{1/5}}{3}\cdot dy=(\frac{5\cdot 32^{6/5}}{18})-(\frac{5}{18})=17.5$$

Which is discrepant with given solution of 157.5

...
1. If $y = 32$ then $x = \frac23$.

2. The area you are looking for must be inside a rectangle (coloured green) with the length 31 and the width $\frac23$. (see attachment)

3. This rectangle has an area of $\frac{62}3 \approx 20.67$.

So the given solution must be wrong.

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#### MarkFL

Staff member
Re: integration of area between curve and y-axis: transpostion question/problem

I think this is a poorly worded problem, and the intended result is to find the area bounded by the given curve, the two given lines and the y-axis.

I therefore agree with the OP's result of 17.5 square units.

I also agree that the stated solution of 157.5 is a typo as described by the OP.

Just my $0.02. #### DeusAbscondus ##### Active member Re: integration of area between curve and y-axis: transpostion question/problem 1. If$y = 32$then$x = \frac23$. 2. The area you are looking for must be inside a rectangle (coloured green) with the length 31 and the width$\frac23$. (see attachment) 3. This rectangle has an area of$\frac{62}3 \approx 20.67$. So the given solution must be wrong. Instructive and encouraging. Thank you, D'Abs #### DeusAbscondus ##### Active member Re: integration of area between curve and y-axis: transpostion question/problem I think this is a poorly worded problem, and the intended result is to find the area bounded by the given curve, the two given lines and the y-axis. I therefore agree with the OP's result of 17.5 square units. I also agree that the stated solution of 157.5 is a typo as described by the OP. Just my$0.02. Encouraging, affirming.
(The course book is riddled with errors and "poorly worded problems", definitions and examples. Fortunately, I'm not without other resources, of which MHB is a bracing, arching, under-girding framework)
Deus Abs
(OP)