Express the volume bounded below by the surface $z=x^2 + y^2$, and above by the plane $z = 2x + 6y + 1$ as the integral of a function over unit ball

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I was trying to solve this question, and calculate the volume of that region, and before reading the answer, I have come up with the following solution, but is my solution correct ?



Solution:



Let $g(x,y) = x^2 + 2y^2$ and $f(x,y) = 2x + 6y + 1$, and we want to take the integral of $1$ for all $z$ satisfying
$$g(x,y) leq z leq f(x,y)$$
for all $x,y$ satisfying the condition.



By the theorem 14.4 in Munkres book (Analysis on Manifolds), we do know that we can express this integral as



$$int_A int_z=g(x,y)^z = f(x,y),$$
where $A = (x,y) in mathbbR^n $.



Now, we have



$$int_A (f-g),$$
where $f-g = 7/2 - (x-1)^2 - (2y - 3/2)^2$, so let $h(u,v) = (u-1, 2v -3/2)$, hence by change of variable theorem we have



$$int_u^2 + y^2 leq 7/2 (7/2 - u^2 - v^2) cdot 2,$$



and we can compute this integral again by applying polar coordinate transformation, and we get
$$= 49/4$$



Is my solution and the reasoning correct ?




real-analysis definite-integrals change-of-variable diffeomorphism




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  • Check this $f-g=-(x-1)^2-2left(y-frac32right)^2+frac132$
    – Lozenges
    Jul 27 at 13:03














up vote
0
down vote

favorite












I was trying to solve this question, and calculate the volume of that region, and before reading the answer, I have come up with the following solution, but is my solution correct ?



Solution:



Let $g(x,y) = x^2 + 2y^2$ and $f(x,y) = 2x + 6y + 1$, and we want to take the integral of $1$ for all $z$ satisfying
$$g(x,y) leq z leq f(x,y)$$
for all $x,y$ satisfying the condition.



By the theorem 14.4 in Munkres book (Analysis on Manifolds), we do know that we can express this integral as



$$int_A int_z=g(x,y)^z = f(x,y),$$
where $A = (x,y) in mathbbR^n $.



Now, we have



$$int_A (f-g),$$
where $f-g = 7/2 - (x-1)^2 - (2y - 3/2)^2$, so let $h(u,v) = (u-1, 2v -3/2)$, hence by change of variable theorem we have



$$int_u^2 + y^2 leq 7/2 (7/2 - u^2 - v^2) cdot 2,$$



and we can compute this integral again by applying polar coordinate transformation, and we get
$$= 49/4$$



Is my solution and the reasoning correct ?




real-analysis definite-integrals change-of-variable diffeomorphism




share|cite|improve this question





















  • Check this $f-g=-(x-1)^2-2left(y-frac32right)^2+frac132$
    – Lozenges
    Jul 27 at 13:03












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I was trying to solve this question, and calculate the volume of that region, and before reading the answer, I have come up with the following solution, but is my solution correct ?



Solution:



Let $g(x,y) = x^2 + 2y^2$ and $f(x,y) = 2x + 6y + 1$, and we want to take the integral of $1$ for all $z$ satisfying
$$g(x,y) leq z leq f(x,y)$$
for all $x,y$ satisfying the condition.



By the theorem 14.4 in Munkres book (Analysis on Manifolds), we do know that we can express this integral as



$$int_A int_z=g(x,y)^z = f(x,y),$$
where $A = (x,y) in mathbbR^n $.



Now, we have



$$int_A (f-g),$$
where $f-g = 7/2 - (x-1)^2 - (2y - 3/2)^2$, so let $h(u,v) = (u-1, 2v -3/2)$, hence by change of variable theorem we have



$$int_u^2 + y^2 leq 7/2 (7/2 - u^2 - v^2) cdot 2,$$



and we can compute this integral again by applying polar coordinate transformation, and we get
$$= 49/4$$



Is my solution and the reasoning correct ?




real-analysis definite-integrals change-of-variable diffeomorphism




share|cite|improve this question













I was trying to solve this question, and calculate the volume of that region, and before reading the answer, I have come up with the following solution, but is my solution correct ?



Solution:



Let $g(x,y) = x^2 + 2y^2$ and $f(x,y) = 2x + 6y + 1$, and we want to take the integral of $1$ for all $z$ satisfying
$$g(x,y) leq z leq f(x,y)$$
for all $x,y$ satisfying the condition.



By the theorem 14.4 in Munkres book (Analysis on Manifolds), we do know that we can express this integral as



$$int_A int_z=g(x,y)^z = f(x,y),$$
where $A = (x,y) in mathbbR^n $.



Now, we have



$$int_A (f-g),$$
where $f-g = 7/2 - (x-1)^2 - (2y - 3/2)^2$, so let $h(u,v) = (u-1, 2v -3/2)$, hence by change of variable theorem we have



$$int_u^2 + y^2 leq 7/2 (7/2 - u^2 - v^2) cdot 2,$$



and we can compute this integral again by applying polar coordinate transformation, and we get
$$= 49/4$$



Is my solution and the reasoning correct ?





real-analysis definite-integrals change-of-variable diffeomorphism





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 27 at 10:25
























asked Jul 27 at 10:08









onurcanbektas

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  • Check this $f-g=-(x-1)^2-2left(y-frac32right)^2+frac132$
    – Lozenges
    Jul 27 at 13:03
















  • Check this $f-g=-(x-1)^2-2left(y-frac32right)^2+frac132$
    – Lozenges
    Jul 27 at 13:03















Check this $f-g=-(x-1)^2-2left(y-frac32right)^2+frac132$
– Lozenges
Jul 27 at 13:03




Check this $f-g=-(x-1)^2-2left(y-frac32right)^2+frac132$
– Lozenges
Jul 27 at 13:03















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