What is the value of the double integral?

[erin]

Hi,

I need to evaluate the following double integral. I have tried direct integration but the answer is too complicated for it to be a viable method.

First integral is from 0 to (1-y^2) function is (x^2+y^2)dx.

Second integral is from 0 to 1 dy.

I can't figure out how to use the maths thing on the right.

Cheers,
Erin

 

[Prove It]

Hi,

I need to evaluate the following double integral. I have tried direct integration but the answer is too complicated for it to be a viable method.

First integral is from 0 to (1-y^2) function is (x^2+y^2)dx.

Second integral is from 0 to 1 dy.

I can't figure out how to use the maths thing on the right.

Cheers,
Erin

Direct integration works, why do you think it doesn't?

 

[erin]

Well obviously if I knew how to do it I wouldn't have posted. Your answer was very helpful. Thank you. I wish I was as smart as you.

 

[I like Serena]

Hi,

I need to evaluate the following double integral. I have tried direct integration but the answer is too complicated for it to be a viable method.

First integral is from 0 to (1-y^2) function is (x^2+y^2)dx.

Second integral is from 0 to 1 dy.

I can't figure out how to use the maths thing on the right.

Cheers,
Erin

Hi Erin!

If I understand correctly, you want to evaluate:
$$\int_0^{1-y^2} (x^2+y^2)dx$$
It can be rewritten as:
$$\int_0^{1-y^2} x^2 dx + \int_0^{1-y^2} y^2 dx$$

How far can you get integrating:
$$\int x^2 dx$$
?

 

[chisigma]

Hi,

I need to evaluate the following double integral. I have tried direct integration but the answer is too complicated for it to be a viable method.

First integral is from 0 to (1-y^2) function is (x^2+y^2)dx.

Second integral is from 0 to 1 dy.

I can't figure out how to use the maths thing on the right.

Cheers,
Erin

Wellcome on MHB erin!...

... using the appropriate notation the integral You want to compute is...

$\displaystyle \int_{0}^{1} \int_{0}^{1 - y^{2}} (x^{2} + y^{2})\ dx\ dy\ (1)$

... isn't it?...

Kind regards

$\chi$ $\sigma$

 

[erin]

Hi chisigma,

yes that is the integral I want to evaluate. Thanks for writing it out properly - are there instructions for how to use the maths symbols somewhere on this forum? I can't find them.

Cheers,
Erin

 

[chisigma]

Hi chisigma,

yes that is the integral I want to evaluate. Thanks for writing it out properly - are there instructions for how to use the maths symbols somewhere on this forum? I can't find them.

Cheers,
Erin

You can find aqn excellent tutorial om Latx in...

mathhelpboards.com/latex-help-discussion-26/

... regarding the integral You have first to solve the 'inner integral'...

$\displaystyle f(y) = \int_{0}^{1 - y^{2}} (x^{+2} + y^{2})\ d x\ (1)$

... and then solve the 'outer integral'...

$\displaystyle I= \int_{0}^{1} f(y)\ d y\ (2)$

... are You able to do that?...

Kind regards

$\chi$ $\sigma$

 

[erin]

You can find aqn excellent tutorial om Latx in...

mathhelpboards.com/latex-help-discussion-26/

So sorry, the upper limit is the square root, not just 1-y^2.

$\displaystyle f(y) = \int_{0}^{\sqrt{1 - y^{2}}} (x^{+2} + y^{2})\ d x\ (1)$

When doing the inner integral I get:
$\displaystyle I= \int_{0}^{1} ((1/3*(1-y^{2})^{3/2}+\sqrt{(1-y^{2})}*y^{2}) d y\ (1)$

I am having difficulty integrating this part with respect to y.

 

[Prove It]

You can find aqn excellent tutorial om Latx in...

mathhelpboards.com/latex-help-discussion-26/

So sorry, the upper limit is the square root, not just 1-y^2.

$\displaystyle f(y) = \int_{0}^{\sqrt{1 - y^{2}}} (x^{+2} + y^{2})\ d x\ (1)$

When doing the inner integral I get:
$\displaystyle I= \int_{0}^{1} ((1/3*(1-y^{2})^{3/2}+\sqrt{(1-y^{2})}*y^{2}) d y\ (1)$

I am having difficulty integrating this part with respect to y.

$\displaystyle \begin{align*} \int_0^1{ \frac{\left( \sqrt{1 - y^2} \right) ^3}{3} + y^2\,\sqrt{1 - y^2}\,\mathrm{d}y } &= \int_0^1{ \frac{\left( 1 - y^2 \right) \sqrt{1 - y^2} }{3} + y^2 \,\sqrt{1 - y^2}\,\mathrm{d}y } \\ &= \frac{1}{3} \int_0^1{ \left( 1 - y^2 \right) \sqrt{1 - y^2} + 3y^2 \,\sqrt{1 - y^2}\,\mathrm{d}y } \\ &= \frac{1}{3} \int_0^1{ \sqrt{1 - y^2} - y^2\,\sqrt{1 - y^2} + 3y^2\,\sqrt{1 - y^2} \,\mathrm{d}y } \\ &= \int_0^1{ \sqrt{1-y^2} + 2y^2\,\sqrt{1-y^2}\,\mathrm{d}y } \end{align*}$

Now to continue you will need to make the substitution $\displaystyle \begin{align*} y = \sin{(\theta )} \implies \mathrm{d}y = \cos{(\theta )} \,\mathrm{d}\theta \end{align*}$.